# IBDP Maths analysis and approaches Topic: AHL 5.16- Integration by substitution HL Paper 1

### Question

(a) Calculate $$\int \frac{1}{5+x^{2}}dx$$ , by using the substitution $$x=\sqrt{5}\, \textup{tan}\, \theta$$
(b) Calculate  (i) $$\int \frac{1}{\sqrt{5-x^{2}}}dx$$ ,   (ii) $$\int \sqrt{5-x^{2}\, dx}$$   by using the substitution $$x=\sqrt{5}\, \textup{sin}\, \theta$$

Ans
(a) $$\frac{1}{\sqrt{5}}\textup{arctan}\frac{x}{\sqrt{5}}+c$$         (b)  (i) $$\textup{arcsin}\frac{x}{\sqrt{5}}+c$$ ,      (ii) $$\frac{5}{2}\textup{arcsin}\frac{x}{\sqrt{5}}+\frac{x\sqrt{5-x^{2}}}{2}+c$$

### Question

Calculate $$\int \sqrt{\frac{x}{4-x}}dx$$ , by using the substitution $$x=4\, \textup{sin}^{2}\, \theta$$
Express your answer in the form $$A\, \textup{arcsin}\frac{\sqrt{x}}{2}+B\sqrt{4x-x^{2}}+c$$

Ans
$$4\, \textup{arcsin}\frac{\sqrt{x}}{2}-\sqrt{4x-x^{2}}+c$$

### Question

(a) Show that $$\frac{2x+4}{(x^{2}+4)(x-2)}=\frac{1}{x-2}-\frac{x}{x^{2}+4}$$
(b) Hence find $$\int \frac{2x+4}{(x^{2}+4)(x-2)}dx$$

Ans
(b) $$\textup{In}(x-2)-\frac{1}{2}\textup{In}(x^{2}+4)+c$$

### Question

Using the substitution $$y=2-x$$, or otherwise, find $$\int \left ( \frac{x}{2-x} \right )^{2}dx$$.                                               (Total 6 marks)

Ans
$$I=\int \left ( \frac{2-y}{y} \right )^{2}(-dy)$$                                                                                          (M1)(A1)
$$=\int \left ( \frac{4}{y^{2}}-\frac{4}{y}+1 \right )\textup{d}y=\frac{4}{y}+4\, \textup{In}\left | y \right |-y+c$$                               (A1)(A1)(A1)
$$=\frac{4}{2-x}+4\textup{In}\left | 2-x \right |-(2-x)+c$$                                                       (A1) (C6)
Note: c and modulus signs not required.                                                                                                               [6]

### Question

Using the substitution  $$u=\frac{1}{2}x+1$$,or otherwise, find the integral $$\int x\sqrt{\frac{1}{2}x+1}\: dx$$.                                               (Total 4 marks)

Ans
Let $$u=\frac{1}{2}x+1\Leftrightarrow x=2(u-1)\Rightarrow \frac{\textup{d}x}{\textup{d}u}=2$$
Then $$\int x\left ( \frac{1}{2}x+1 \right )^{1/2}dx$$
$$=\int 2(u-1)\times u^{1/2}\times 2\textup{d}u$$                                                        (M1)
$$=4\int (u^{3/2}-u^{1/2})\textup{d}u$$                                                                      (A1)
$$=4\left [ \frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2} \right ]+C$$                                                          (M1)
$$=4\left [ \frac{2}{5}\left ( \frac{1}{2}x+1 \right )^{5/2}-\frac{2}{3}\left ( \frac{1}{2}x+1 \right )^{3/2} \right ]+C$$                 (A1)                                 [4]

### Question

Use the substitution $$u=x+2$$ to find $$\int \frac{x^{3}}{(x+2)^{2}}\textup{d}x$$.                                           (Total 6 marks)

Ans
Substituting $$u=x+2\Rightarrow u-2=x,\textup{d}u=\textup{d}x$$                      (M1)
$$\int \frac{x^{3}}{(x+2)^{2}}\textup{d}x=\frac{(u-2)^{3}}{u^{2}}\textup{d}u$$                                                                             A1
$$=\int \frac{u^{3}-6u^{2}+12u-8}{u^{2}}\textup{d}u$$                                                                                    A1
$$\int u\, \textup{d}u+\int (-6)\textup{d}u+\int \frac{12}{u}\textup{d}u+\int 8u^{-2}\textup{d}u$$                                A1
$$=\frac{u^{2}}{2}-6u+12\textup{In}\left | u \right |+8u^{-1}+c$$                                                   A1
$$=\frac{(x+2)^{2}}{2}-6(x+2)+12\, \textup{In}\left | x+2 \right |+\frac{8}{x+2}+c$$                  A1                                 [6]

### Question

Find $$\int \frac{e^{x}}{e^{2x}+9}dx$$.                                                     (Total 5 marks)
Extra question
Find in a similar way the integral $$\int \frac{3^{x}}{9^{x}+9}dx$$

Ans
$$\textup{Let }u=e^{x}$$                                                                                                                                       M1
$$\textup{d}u=e^{x}\textup{d}x\, (\textup{or equivalent})$$                                                                                                A1
$$\textup{When}\, x=0,u=1\, \textbf{and}\, \textup{when}\, x=\textup{In}\, 3, u=3$$                                                 (A1)
$$\int_{0}^{\textup{In}3}\frac{e^{x}}{e^{2x}+9}dx=\int_{1}^{3}\frac{1}{u^{2}+9}\textup{d}u$$                                                                                                 A1
$$=\frac{1}{3}\left [ \textup{arctan}\frac{u}{3} \right ]^{3}_{1}$$                                                                                                                          A1
$$=\frac{1}{3}\left ( \textup{arctan}1-\textup{arctan}\frac{1}{3} \right )\left ( =\frac{\pi }{12}-\frac{1}{3}\textup{arctan}\frac{1}{3},\frac{1}{3}\textup{arctan}\frac{1}{2} \right )$$                 A1              N0                  [6]
Extra question
Find in a similar way the integral $$\int \frac{3^{x}}{9^{x}+9}dx$$

### Question

By using an appropriate substitution find $$\int \frac{\textup{tan}(\textup{In}\, y)}{y}dy,y> 0$$.                   (Total 6 marks)
Extra question
The integral $$\int \frac{\textup{cot}(\textup{In}\, y)}{y}dy,y> 0$$ can be found in a similar way. Write down the result!

Ans
$$\textup{Let}\, u=\textup{In}\, y\Rightarrow \textup{d}u=\frac{1}{y}\textup{d}y$$                                 A1(A1)
$$\int \frac{\textup{tan}(\textup{In}\, y)}{y}\textup{d}y=\int \textup{tan}\, u\, \textup{d}u$$                                   A1
$$\int \frac{\textup{sin}\, u}{\textup{cos}\, u}\textup{d}u=-\textup{In}\left | \textup{cos}\, u \right |+c$$                              A1
EITHER
$$\int \frac{\textup{tan}(\textup{In}\, y)}{y}\textup{d}y=-\textup{In}\left | \textup{cos}(\textup{In}\, y) \right |+c$$             A1A1
OR
$$\int \frac{\textup{tan}(\textup{In}\, y)}{y}\textup{d}y=\textup{In}\left | \textup{sec}(\textup{In}\, y) \right |+c$$                 A1A1                                                                         [6]

Extra question
$$\textup{In}\left | \textup{sin}(\textup{In}\, y) \right |+c$$

### Question

Using the substitution $$2x=\textup{sin}\theta$$, or otherwise, find $$\int \left ( \sqrt{1-4x^{2}} \right )dx$$.                                                  (Total 6 marks)

Ans
$$\int \left (\sqrt{1-4x^{2}} \right )\textup{d}x$$
$$\textup{Let}\, 2x=\textup{sin}\theta \Rightarrow 2\frac{\textup{d}x}{\textup{d}\theta }=\textup{cos}\, \theta \Rightarrow \textup{d}x=\frac{1}{2}\textup{cos}\, \theta \, \textup{d}\, \theta$$
$$\Rightarrow \int \left ( \sqrt{1-4x^{2}} \right )\textup{d}x=\int \sqrt{1-\textup{sin}^{2}\theta \frac{1}{2}}\textup{cos}\,\theta \textup{d}\, \theta$$
$$=\int \frac{1}{2}\textup{cos}^{2}\theta \, \textup{d}\, \theta$$                                                                                      (A1)
$$\int \frac{1}{4}(\textup{cos}\, 2\theta +1)\textup{d}\theta$$                                                                              (A1)
$$=\frac{1}{8}\textup{sin}\, 2\theta +\frac{\theta }{4}+C$$                                                                          (A1)(A1)
$$=\frac{1}{4}\left [ 2x\sqrt{1-4x^{2}}+\textup{arcsin}\, 2x \right ]+C$$                                  (A1)(A1)    (C6)                                        [6]

### Question: [Maximum mark: 21]

The function f is defined by f (x) = ex sinx , where x ∈ R.

(a) Find the Maclaurin series for f (x) up to and including the x3 term.
(b) Hence, find an approximate value for $$\int_{0}^{1}e^{x^{2}} sin(x^{2})dx.$$

The function g is defined by g (x) = ex cos x , where x ∈ R.

(c) (i) Show that g(x) satisfies the equation g ″(x) = 2(g′(x) – g(x)).
(ii) Hence, deduce that g(4) (x) = 2(g″′(x) – g ″(x)) .

(d) Using the result from part (c), find the Maclaurin series for g(x) up to and including the x4 term.

(e) Hence, or otherwise, determine the value of $$\lim_{x\rightarrow 0}\frac{e^{x}cos x – 1 – x}{x^{3}}.$$

Ans:

(a) METHOD 1
recognition of both known series

attempt to multiply the two series up to and including x3 term

Note: Condone absence of limits up to this stage.

Note: Accept working with each side separately to obtain -2ex sin x .

Note: Accept working with each side separately to obtain -4ex cos x .

Note: Do not award any marks for approaches that do not use the part (c) result.

Note: Condone the omission of +… in their working.

### Question

By using the substitution x = tan u , find the value of $$\int_0 ^1 \frac{X^2}{(1+X^2)^3}$$ [Maximum mark: 8]

Ans:

x= tanu $$\Rightarrow \frac{dx}{du}= sec^{2}u$$ OR u= arctan x $$\Rightarrow \frac{du}{dx}$$= $$\frac{1}{1+x^{2}}$$attempt to write the integral in terms of u  $$\int_{0}^{\frac{\pi }{4}}\frac{tan^{2}u sec^{2}udu}{(1+tan^{2}u)^{3}} = \int_{0}^{\frac{\pi }{4}}\frac{tan^{2}u sec^{2}udu}{(sec^{2}u)^{3}}= \int_{0}^{\frac{\pi }{4}}sin^{2} u cos^{2}du = \frac{1}{4}\int_{0}^{\frac{\pi }{4}}sin^{2}2udu= \frac{1}{8}\int_{0}^{\frac{\pi }{4}}(1-cos4u)du$$

### Question

By using the substitution u = sin x , find $$\int \frac{sinx. cosx}{sin^2x-sinx-2}dx$$

Ans:

Let $u=\sin x$, then $\frac{\text{d}u}{\text{d}x}=\cos x$.
$$\begin{eqnarray} \int\frac{\sin x\cos x}{\sin ^2-\sin x-2} \text{d}x &=& \int\frac{u}{u^2-u-2} \text{d}u \nonumber \\ &=& \int\frac{u}{\left(u-2\right)\left(u+1\right)} \text{d}u \nonumber \\ &=& \frac{1}{3}\int\frac{2}{u-2}+\frac{1}{u+1} \text{d}u \nonumber \\ &=& \frac{1}{3}\left[2\ln\left|u-2\right|+\ln\left|u+1\right|\right]+c \nonumber \\ &=& \frac{1}{3}\left[\ln\left|\left(u-2\right)^2\left(u+1\right)\right|\right]+c \nonumber \\ &=& \frac{1}{3}\left[\ln\left|\left(\sin x-2\right)^2\left(\sin x+1\right)\right|\right]+c. \end{eqnarray}$$

### Question

By using the substitution u = sin x , find $$\int \frac{sinx. cosx}{sin^2x-sinx-2}dx$$

Ans:

Let $u=\sin x$, then $\frac{\text{d}u}{\text{d}x}=\cos x$.
$$\begin{eqnarray} \int\frac{\sin x\cos x}{\sin ^2-\sin x-2} \text{d}x &=& \int\frac{u}{u^2-u-2} \text{d}u \nonumber \\ &=& \int\frac{u}{\left(u-2\right)\left(u+1\right)} \text{d}u \nonumber \\ &=& \frac{1}{3}\int\frac{2}{u-2}+\frac{1}{u+1} \text{d}u \nonumber \\ &=& \frac{1}{3}\left[2\ln\left|u-2\right|+\ln\left|u+1\right|\right]+c \nonumber \\ &=& \frac{1}{3}\left[\ln\left|\left(u-2\right)^2\left(u+1\right)\right|\right]+c \nonumber \\ &=& \frac{1}{3}\left[\ln\left|\left(\sin x-2\right)^2\left(\sin x+1\right)\right|\right]+c. \end{eqnarray}$$

### Question

Calculate $$\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x}$$ .[6]

a.

Find $$\int {{{\tan }^3}x{\text{d}}x}$$ .[3]

b.

### Markscheme

EITHER

let $$u = \tan x;{\text{ d}}u = {\sec ^2}x{\text{d}}x$$     (M1)

consideration of change of limits     (M1)

$$\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} = \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{1}{{{u^{\frac{1}{3}}}}}{\text{d}}u}$$     (A1)

Note: Do not penalize lack of limits.

$$= \left[ {\frac{{3{u^{\frac{2}{3}}}}}{2}} \right]_1^{\sqrt 3 }$$     A1

$$= \frac{{3 \times {{\sqrt 3 }^{\frac{2}{3}}}}}{2} – \frac{3}{2} = \left( {\frac{{3\sqrt[3]{3} – 3}}{2}} \right)$$     A1A1     N0

OR

$$\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} = \left[ {\frac{{3{{(\tan x)}^{\frac{2}{3}}}}}{2}} \right]_{\frac{\pi }{4}}^{\frac{\pi }{3}}$$     M2A2

$$= \frac{{3 \times {{\sqrt 3 }^{\frac{2}{3}}}}}{2} – \frac{3}{2} = \left( {\frac{{3\sqrt[3]{3} – 3}}{2}} \right)$$     A1A1     N0

[6 marks]

a.

$$\int {{{\tan }^3}x{\text{d}}x} = \int {\tan x({{\sec }^2}x – 1){\text{d}}x}$$     M1

$$= \int {(\tan x \times {{\sec }^2}x – \tan x){\text{d}}x}$$

$$= \frac{1}{2}{\tan ^2}x – \ln \left| {\sec x} \right| + C$$     A1A1

Note: Do not penalize the absence of absolute value or C.

[3 marks]

b.

### Question

(a)     Given that $$\alpha > 1$$, use the substitution $$u = \frac{1}{x}$$ to show that

$\int_1^\alpha {\frac{1}{{1 + {x^2}}}{\text{d}}x = \int_{\frac{1}{\alpha }}^1 {\frac{1}{{1 + {u^2}}}{\text{d}}x} } .$

(b)     Hence show that $$\arctan \alpha + \arctan \frac{1}{\alpha } = \frac{\pi }{2}$$.

### Markscheme

(a)     $$u = \frac{1}{x} \Rightarrow {\text{d}}u = – \frac{1}{{{x^2}}}{\text{d}}x$$     M1

$$\Rightarrow {\text{d}}x = – \frac{{{\text{d}}u}}{{{u^2}}}$$     A1

$$\int_1^\alpha {\frac{1}{{1 + {x^2}}}{\text{d}}x = – \int_1^{\frac{1}{\alpha }} {\frac{1}{{1 + {{\left( {\frac{1}{u}} \right)}^2}}}\frac{{{\text{d}}u}}{{{u^2}}}} }$$     A1M1A1

Note: Award A1 for correct integrand and M1A1 for correct limits.

$$= \int_{\frac{1}{\alpha }}^1 {\frac{1}{{1 + {u^2}}}{\text{d}}u\,\,\,\,\,}$$(upon interchanging the two limits)     AG

(b)     $$\arctan x_1^\alpha = \arctan u_{\frac{1}{\alpha }}^1$$     A1

$$\arctan \alpha – \frac{\pi }{4} = \frac{\pi }{4} – \arctan \frac{1}{\alpha }$$     A1

$$\arctan \alpha + \arctan \frac{1}{\alpha } = \frac{\pi }{2}$$     AG

[7 marks]

### Question

(i)     Sketch the graphs of $$y = \sin x$$ and $$y = \sin 2x$$ , on the same set of axes, for $$0 \leqslant x \leqslant \frac{\pi }{2}$$ .

(ii)     Find the x-coordinates of the points of intersection of the graphs in the domain $$0 \leqslant x \leqslant \frac{\pi }{2}$$ .

(iii)     Find the area enclosed by the graphs.[9]

a.

Find the value of $$\int_0^1 {\sqrt {\frac{x}{{4 – x}}} }{{\text{d}}x}$$ using the substitution $$x = 4{\sin ^2}\theta$$ .[8]

b.

The increasing function f satisfies $$f(0) = 0$$ and $$f(a) = b$$ , where $$a > 0$$ and $$b > 0$$ .

(i)     By reference to a sketch, show that $$\int_0^a {f(x){\text{d}}x = ab – \int_0^b {{f^{ – 1}}(x){\text{d}}x} }$$ .

(ii)     Hence find the value of $$\int_0^2 {\arcsin \left( {\frac{x}{4}} \right){\text{d}}x}$$ .[8]

c.

### Markscheme

(i)

A2

Note: Award A1 for correct $$\sin x$$ , A1 for correct $$\sin 2x$$ .

Note: Award A1A0 for two correct shapes with $$\frac{\pi }{2}$$ and/or 1 missing.

Note: Condone graph outside the domain.

(ii)     $$\sin 2x = \sin x$$ , $$0 \leqslant x \leqslant \frac{\pi }{2}$$

$$2\sin x\cos x – \sin x = 0$$     M1

$$\sin x(2\cos x – 1) = 0$$

$$x = 0,\frac{\pi }{3}$$     A1A1     N1N1

(iii)     area $$= \int_0^{\frac{\pi }{3}} {(\sin 2x – \sin x){\text{d}}x}$$     M1

Note: Award M1 for an integral that contains limits, not necessarily correct, with $$\sin x$$ and $$\sin 2x$$ subtracted in either order.

$$= \left[ { – \frac{1}{2}\cos 2x + \cos x} \right]_0^{\frac{\pi }{3}}$$     A1

$$= \left( { – \frac{1}{2}\cos \frac{{2\pi }}{3} + \cos \frac{\pi }{3}} \right) – \left( { – \frac{1}{2}\cos 0 + \cos 0} \right)$$     (M1)

$$= \frac{3}{4} – \frac{1}{2}$$

$$= \frac{1}{4}$$     A1

[9 marks]

a.

$$\int_0^1 {\sqrt {\frac{x}{{4 – x}}} } {\text{d}}x = \int_0^{\frac{\pi }{6}} {\sqrt {\frac{{4{{\sin }^2}\theta }}{{4 – 4{{\sin }^2}\theta }}} \times 8\sin \theta \cos \theta {\text{d}}\theta }$$     M1A1A1

Note: Award M1 for substitution and reasonable attempt at finding expression for dx in terms of $${\text{d}}\theta$$ , first A1 for correct limits, second A1 for correct substitution for dx .

$$\int_0^{\frac{\pi }{6}} {8{{\sin }^2}\theta {\text{d}}\theta }$$     A1

$$\int_0^{\frac{\pi }{6}} {4 – 4\cos 2\theta {\text{d}}\theta }$$     M1

$$= [4\theta – 2\sin 2\theta ]_0^{\frac{\pi }{6}}$$     A1

$$= \left( {\frac{{2\pi }}{3} – 2\sin \frac{\pi }{3}} \right) – 0$$     (M1)

$$= \frac{{2\pi }}{3} – \sqrt 3$$     A1

[8 marks]

b.

(i)

M1

from the diagram above

the shaded area $$= \int_0^a {f(x){\text{d}}x = ab – \int_0^b {{f^{ – 1}}(y){\text{d}}y} }$$     R1

$${ = ab – \int_0^b {{f^{ – 1}}(x){\text{d}}x} }$$     AG

(ii)     $$f(x) = \arcsin \frac{x}{4} \Rightarrow {f^{ – 1}}(x) = 4\sin x$$     A1

$$\int_0^2 {\arcsin \left( {\frac{x}{4}} \right){\text{d}}x = \frac{\pi }{3} – \int_0^{\frac{\pi }{6}} {4\sin x{\text{d}}x} }$$     M1A1A1

Note: Award A1 for the limit $$\frac{\pi }{6}$$ seen anywhere, A1 for all else correct.

$$= \frac{\pi }{3} – [ – 4\cos x]_0^{\frac{\pi }{6}}$$     A1

$$= \frac{\pi }{3} – 4 + 2\sqrt 3$$     A1

Note: Award no marks for methods using integration by parts.

[8 marks]

c.

### Question

Let $$f(x) = \sqrt {\frac{x}{{1 – x}}} ,{\text{ }}0 < x < 1$$.

Show that $$f'(x) = \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}$$ and deduce that f is an increasing function.[5]

a.

Show that the curve $$y = f(x)$$ has one point of inflexion, and find its coordinates.[6]

b.

Use the substitution $$x = {\sin ^2}\theta$$ to show that $$\int {f(x){\text{d}}x} = \arcsin \sqrt x – \sqrt {x – {x^2}} + c$$ .[11]

c.

## Markscheme

EITHER

derivative of $$\frac{x}{{1 – x}}$$ is $$\frac{{(1 – x) – x( – 1)}}{{{{(1 – x)}^2}}}$$     M1A1

$$f'(x) = \frac{1}{2}{\left( {\frac{x}{{1 – x}}} \right)^{ – \frac{1}{2}}}\frac{1}{{{{(1 – x)}^2}}}$$     M1A1

$$= \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}$$     AG

$$f'(x) > 0$$ (for all $$0 < x < 1$$) so the function is increasing     R1

OR

$$f(x) = \frac{{{x^{\frac{1}{2}}}}}{{{{(1 – x)}^{\frac{1}{2}}}}}$$

$$f'(x) = \frac{{{{(1 – x)}^{\frac{1}{2}}}\left( {\frac{1}{2}{x^{ – \frac{1}{2}}}} \right) – \frac{1}{2}{x^{\frac{1}{2}}}{{(1 – x)}^{ – \frac{1}{2}}}( – 1)}}{{1 – x}}$$     M1A1

$$= \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{1}{2}}} + \frac{1}{2}{x^{\frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}$$     A1

$$= \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}[1 – x + x]$$     M1

$$= \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}$$     AG

$$f'(x) > 0$$ (for all $$0 < x < 1$$) so the function is increasing     R1

[5 marks]

a.

$$f'(x) = \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}$$

$$\Rightarrow f”(x) = -\frac{1}{4}{x^{ – \frac{3}{2}}}{(1 – x)^{ – \frac{3}{2}}} + \frac{3}{4}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{5}{2}}}$$     M1A1

$$= -\frac{1}{4}{x^{ – \frac{3}{2}}}{(1 – x)^{ – \frac{5}{2}}}[1 – 4x]$$

$$f”(x) = 0 \Rightarrow x = \frac{1}{4}$$     M1A1

$$f”(x)$$ changes sign at $$x = \frac{1}{4}$$ hence there is a point of inflexion     R1

$$x = \frac{1}{4} \Rightarrow y = \frac{1}{{\sqrt 3 }}$$     A1

the coordinates are $$\left( {\frac{1}{4},\frac{1}{{\sqrt 3 }}} \right)$$

[6 marks]

b.

$$x = {\sin ^2}\theta \Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}\theta }} = 2\sin \theta \cos \theta$$     M1A1

$$\int {\sqrt {\frac{x}{{1 – x}}} {\text{d}}x = \int {\sqrt {\frac{{{{\sin }^2}\theta }}{{1 – {{\sin }^2}\theta }}} 2\sin \theta \cos \theta {\text{d}}\theta } }$$     M1A1

$$= \int {2{{\sin }^2}\theta {\text{d}}\theta }$$     A1

$$= \int {1 – \cos 2\theta } {\text{d}}\theta$$     M1A1

$$= \theta – \frac{1}{2}\sin 2\theta + c$$     A1

$$\theta = \arcsin \sqrt x$$     A1

$$\frac{1}{2}\sin 2\theta = \sin \theta \cos \theta = \sqrt x \sqrt {1 – x} = \sqrt {x – {x^2}}$$     M1A1

hence $$\int {\sqrt {\frac{x}{{1 – x}}} {\text{d}}x = \arcsin \sqrt x } – \sqrt {x – {x^2}} + c$$     AG

[11 marks]

c.

### Question

The function f is defined by $$f(x) = \frac{1}{{4{x^2} – 4x + 5}}$$.

Express $$4{x^2} – 4x + 5$$ in the form $$a{(x – h)^2} + k$$ where a, h, $$k \in \mathbb{Q}$$.[2]

a.

The graph of $$y = {x^2}$$ is transformed onto the graph of $$y = 4{x^2} – 4x + 5$$. Describe a sequence of transformations that does this, making the order of transformations clear.[3]

b.

Sketch the graph of $$y = f(x)$$.[2]

c.

Find the range of f.[2]

d.

By using a suitable substitution show that $$\int {f(x){\text{d}}x = \frac{1}{4}\int {\frac{1}{{{u^2} + 1}}{\text{d}}u} }$$.[3]

e.

Prove that $$\int_1^{3.5} {\frac{1}{{4{x^2} – 4x + 5}}{\text{d}}x = \frac{\pi }{{16}}}$$.[7]

f.

### Markscheme

$$4{(x – 0.5)^2} + 4$$     A1A1

Note: A1 for two correct parameters, A2 for all three correct.

[2 marks]

a.

translation $$\left( {\begin{array}{*{20}{c}} {0.5} \\ 0 \end{array}} \right)$$ (allow “0.5 to the right”)     A1

stretch parallel to y-axis, scale factor 4 (allow vertical stretch or similar)     A1

translation $$\left( {\begin{array}{*{20}{c}} 0 \\ 4 \end{array}} \right)$$ (allow “4 up”)     A1

Note: All transformations must state magnitude and direction.

Note: First two transformations can be in either order.

It could be a stretch followed by a single translation of $$\left( {\begin{array}{*{20}{c}} {0.5} \\ 4 \end{array}} \right)$$. If the vertical translation is before the stretch it is $$\left( {\begin{array}{*{20}{c}} 0 \\ 1 \end{array}} \right)$$.

[3 marks]

b.

general shape (including asymptote and single maximum in first quadrant),     A1

intercept $$\left( {0,\frac{1}{5}} \right)$$ or maximum $$\left( {\frac{1}{2},\frac{1}{4}} \right)$$ shown     A1

[2 marks]

c.

$$0 < f(x) \leqslant \frac{1}{4}$$     A1A1

Note: A1 for $$\leqslant \frac{1}{4}$$, A1 for $$0 <$$.

[2 marks]

d.

let $$u = x – \frac{1}{2}$$     A1

$$\frac{{{\text{d}}u}}{{{\text{d}}x}} = 1\,\,\,\,\,{\text{(or d}}u = {\text{d}}x)$$     A1

$$\int {\frac{1}{{4{x^2} – 4x + 5}}{\text{d}}x = \int {\frac{1}{{4{{\left( {x – \frac{1}{2}} \right)}^2} + 4}}{\text{d}}x} }$$     A1

$$\int {\frac{1}{{4{u^2} + 4}}{\text{d}}u = \frac{1}{4}\int {\frac{1}{{{u^2} + 1}}{\text{d}}u} }$$     AG

Note: If following through an incorrect answer to part (a), do not award final A1 mark.

[3 marks]

e.

$$\int_1^{3.5} {\frac{1}{{4{x^2} – 4x + 5}}{\text{d}}x = \frac{1}{4}\int_{0.5}^3 {\frac{1}{{{u^2} + 1}}{\text{d}}u} }$$     A1

Note: A1 for correct change of limits. Award also if they do not change limits but go back to x values when substituting the limit (even if there is an error in the integral).

$$\frac{1}{4}\left[ {\arctan (u)} \right]_{0.5}^3$$     (M1)

$$\frac{1}{4}\left( {\arctan (3) – \arctan \left( {\frac{1}{2}} \right)} \right)$$     A1

let the integral = I

$$\tan 4I = \tan \left( {\arctan (3) – \arctan \left( {\frac{1}{2}} \right)} \right)$$     M1

$$\frac{{3 – 0.5}}{{1 + 3 \times 0.5}} = \frac{{2.5}}{{2.5}} = 1$$     (M1)A1

$$4I = \frac{\pi }{4} \Rightarrow I = \frac{\pi }{{16}}$$     A1AG

[7 marks]

f.

### Question

Use the substitution $$x = a\sec \theta$$ to show that $$\int_{a\sqrt 2 }^{2a} {\frac{{{\text{d}}x}}{{{x^3}\sqrt {{x^2} – {a^2}} }} = \frac{1}{{24{a^3}}}\left( {3\sqrt 3 + \pi – 6} \right)}$$.

### Markscheme

$$x = a\sec \theta$$

$$\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = a\sec \theta \tan \theta$$     (A1)

new limits:

$$x = a\sqrt 2 \Rightarrow \theta = \frac{\pi }{4}$$ and $$x = 2a \Rightarrow \theta = \frac{\pi }{3}$$     (A1)

$$\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{a\sec \theta \tan \theta }}{{{a^3}{{\sec }^3}\theta \sqrt {{a^2}{{\sec }^2}\theta – {a^2}} }}{\text{d}}\theta }$$     M1

$$= \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\cos }^2}\theta }}{{{a^3}}}{\text{d}}\theta }$$     A1

using $${\cos ^2}\theta = \frac{1}{2}(\cos 2\theta + 1)$$     M1

$$\frac{1}{{2{a^3}}}\left[ {\frac{1}{2}\sin 2\theta + \theta } \right]_{\frac{\pi }{4}}^{\frac{\pi }{3}}$$ or equivalent     A1

$$= \frac{1}{{4{a^3}}}\left( {\frac{{\sqrt 3 }}{2} + \frac{{2\pi }}{3} – 1 – \frac{\pi }{2}} \right)$$ or equivalent     A1

$$= \frac{1}{{24{a^3}}}\left( {3\sqrt 3 + \pi – 6} \right)$$     AG

[7 marks]

### Question

By using the substitution $$u = 1 + \sqrt x$$, find $$\int {\frac{{\sqrt x }}{{1 + \sqrt x }}{\text{d}}x}$$.

### Markscheme

$$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{{2\sqrt x }}$$     A1

$${\text{d}}x = 2(u – 1){\text{d}}u$$

Note:     Award the A1 for any correct relationship between $${\text{d}}x$$ and $${\text{d}}u$$.

$$\int {\frac{{\sqrt x }}{{1 + \sqrt x }}{\text{d}}x} = 2\int {\frac{{{{(u – 1)}^2}}}{u}{\text{d}}u}$$     (M1)A1

Note:     Award the M1 for an attempt at substitution resulting in an integral only involving $$u$$ .

$$= 2\int {u – 2 + \frac{1}{u}{\text{d}}u}$$     (A1)

$$= {u^2} – 4u + 2\ln u( + C)$$     A1

$$= x – 2\sqrt x – 3 + 2\ln \left( {1 + \sqrt x } \right)( + C)$$     A1

Note:     Award the A1 for a correct expression in $$x$$, but not necessarily fully expanded/simplified.

[6 marks]

### Question

By using the substitution $$u = {{\text{e}}^x} + 3$$, find $$\int {\frac{{{{\text{e}}^x}}}{{{{\text{e}}^{2x}} + 6{{\text{e}}^x} + 13}}{\text{d}}x}$$.

### Markscheme

$$\frac{{{\text{d}}u}}{{{\text{d}}x}} = {{\text{e}}^x}$$     (A1)

EITHER

integral is $$\int {\frac{{{{\text{e}}^x}}}{{{{({{\text{e}}^x} + 3)}^2} + {2^2}}}{\text{d}}x}$$     M1A1

$$= \frac{1}{{{u^2} + {2^2}}}{\text{d}}u$$     M1A1

Note:     Award M1 only if the integral has completely changed to one in $$u$$.

Note:     $${\text{d}}u$$ needed for final A1

OR

$${{\text{e}}^x} = u – 3$$

integral is $$\int {\frac{1}{{{{(u – 3)}^2} + 6(u – 3) + 13}}{\text{d}}u}$$     M1A1

Note: Award M1 only if the integral has completely changed to one in $$u$$.

$$= \int {\frac{1}{{{u^2} + {2^2}}}{\text{d}}u}$$     M1A1

Note:     In both solutions the two method marks are independent.

THEN

$$= \frac{1}{2}\arctan \left( {\frac{u}{2}} \right)( + c)$$     (A1)

$$= \frac{1}{2}\arctan \left( {\frac{{{{\text{e}}^x} + 3}}{2}} \right)( + c)$$     A1

Total [7 marks]

### Question

By using the substitution $$t = \tan x$$, find $$\int {\frac{{{\text{d}}x}}{{1 + {{\sin }^2}x}}}$$.

Express your answer in the form $$m\arctan (n\tan x) + c$$, where $$m$$, $$n$$ are constants to be determined.

### Markscheme

EITHER

$$x = \arctan t$$     (M1)

$$\frac{{{\text{d}}x}}{{{\text{d}}t}} = \frac{1}{{1 + {t^2}}}$$     A1

OR

$$t = \tan x$$

$$\frac{{{\text{d}}t}}{{{\text{d}}x}} = {\sec ^2}x$$     (M1)

$$= 1 + {\tan ^2}x$$     A1

$$= 1 + {t^2}$$

THEN

$$\sin x = \frac{t}{{\sqrt {1 + {t^2}} }}$$     (A1)

Note:     This A1 is independent of the first two marks

$$\int {\frac{{{\text{d}}x}}{{1 + {{\sin }^2}x}} = \int {\frac{{\frac{{{\text{d}}t}}{{1 + {t^2}}}}}{{1 + {{\left( {\frac{t}{{\sqrt {1 + {t^2}} }}} \right)}^2}}}} }$$     M1A1

Note:     Award M1 for attempting to obtain integral in terms of $$t$$ and $${\text{d}}t$$

$$= \int {\frac{{{\text{d}}t}}{{(1 + {t^2}) + {t^2}}} = \int {\frac{{{\text{d}}t}}{{1 + 2{t^2}}}} }$$     A1

$$= \frac{1}{2}\int {\frac{{{\text{d}}t}}{{\frac{1}{2} + {t^2}}} = \frac{1}{2} \times \frac{1}{{\frac{1}{{\sqrt 2 }}}}\arctan \left( {\frac{t}{{\frac{1}{{\sqrt 2 }}}}} \right)}$$     A1

$$= \frac{{\sqrt 2 }}{2}\arctan \left( {\sqrt 2 \tan x} \right)( + c)$$     A1

[8 marks]

### Question

Use the substitution $$u = \ln x$$ to find the value of $$\int_{\text{e}}^{{{\text{e}}^2}} {\frac{{{\text{d}}x}}{{x\ln x}}}$$.

### Markscheme

METHOD 1

$$\int_{\text{e}}^{{{\text{e}}^2}} {\frac{{{\text{d}}x}}{{x\ln x}}} = \left[ {\ln (\ln x)} \right]_{\text{e}}^{{{\text{e}}^2}}$$     (M1)A1

$$= \ln (\ln {{\text{e}}^2}) – \ln (\ln {\text{e}})\;\;\;( = \ln 2 – \ln 1)$$     (A1)

$$= \ln 2$$     A1

[4 marks]

METHOD 2

$$u = \ln x,{\text{ }}\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}$$     M1

$$= \int_1^2 {\frac{{{\text{d}}u}}{u}}$$     A1

Note:     Condone absent or incorrect limits here.

$$= [\ln u]_1^2$$ or equivalent in $$x( = \ln 2 – \ln 1)$$     (A1)

$$= \ln 2$$     A1

[4 marks]

### Question

Consider the function defined by $$f(x) = x\sqrt {1 – {x^2}}$$ on the domain $$– 1 \le x \le 1$$.

Show that $$f$$ is an odd function.[2]

a.

Find $$f'(x)$$.[3]

b.

Hence find the $$x$$-coordinates of any local maximum or minimum points.[3]

c.

Find the range of $$f$$.[3]

d.

Sketch the graph of $$y = f(x)$$ indicating clearly the coordinates of the $$x$$-intercepts and any local maximum or minimum points.[3]

e.

Find the area of the region enclosed by the graph of $$y = f(x)$$ and the $$x$$-axis for $$x \ge 0$$.[4]

f.

Show that $$\int_{ – 1}^1 {\left| {x\sqrt {1 – {x^2}} } \right|{\text{d}}x > \left| {\int_{ – 1}^1 {x\sqrt {1 – {x^2}} {\text{d}}x} } \right|}$$.[2]

g.

### Markscheme

$$f( – x) = ( – x)\sqrt {1 – {{( – x)}^2}}$$     M1

$$= – x\sqrt {1 – {x^2}}$$

$$= – f(x)$$     R1

hence $$f$$ is odd     AG

[2 marks]

a.

$$f'(x) = x \bullet \frac{1}{2}{(1 – {x^2})^{ – \frac{1}{2}}} \bullet – 2x + {(1 – {x^2})^{\frac{1}{2}}}$$     M1A1A1

[3 marks]

b.

$$f'(x) = \sqrt {1 – {x^2}} – \frac{{{x^2}}}{{\sqrt {1 – {x^2}} }}\;\;\;\left( { = \frac{{1 – 2{x^2}}}{{\sqrt {1 – {x^2}} }}} \right)$$     A1

Note:     This may be seen in part (b).

Note:     Do not allow FT from part (b).

$$f'(x) = 0 \Rightarrow 1 – 2{x^2} = 0$$     M1

$$x = \pm \frac{1}{{\sqrt 2 }}$$     A1

[3 marks]

c.

$$y$$-coordinates of the Max Min Points are $$y = \pm \frac{1}{2}$$     M1A1

so range of $$f(x)$$ is $$\left[ { – \frac{1}{2},{\text{ }}\frac{1}{2}} \right]$$     A1

Note:     Allow FT from (c) if values of $$x$$, within the domain, are used.

[3 marks]

d.

Shape: The graph of an odd function, on the given domain, s-shaped,

where the max(min) is the right(left) of $$0.5{\text{ }}( – 0.5)$$     A1

$$x$$-intercepts     A1

turning points     A1

[3 marks]

e.

$${\text{area}} = \int_0^1 {x\sqrt {1 – {x^2}} {\text{d}}x}$$     (M1)

attempt at “backwards chain rule” or substitution     M1

$$= – \frac{1}{2}\int_0^1 {( – 2x)\sqrt {1 – {x^2}} {\text{d}}x}$$

Note:     Condone absence of limits for first two marks.

$$= \left[ {\frac{2}{3}{{(1 – {x^2})}^{\frac{3}{2}}} \bullet – \frac{1}{2}} \right]_0^1$$     A1

$$= \left[ { – \frac{1}{3}{{(1 – {x^2})}^{\frac{3}{2}}}} \right]_0^1$$

$$= 0 – \left( { – \frac{1}{3}} \right) = \frac{1}{3}$$     A1

[4 marks]

f.

$$\int_{ – 1}^1 {\left| {x\sqrt {1 – {x^2}} } \right|{\text{d}}x > 0}$$     R1

$$\left| {\int_{ – 1}^1 {x\sqrt {1 – {x^2}} {\text{d}}x} } \right| = 0$$     R1

so $$\int_{ – 1}^1 {\left| {x\sqrt {1 – {x^2}} } \right|{\text{d}}x > \left| {\int_{ – 1}^1 {x\sqrt {1 – {x^2}} {\text{d}}x} } \right|}$$     AG

[2 marks]

Total [20 marks]

g.

### Question

The following diagram shows the graph of $$y = \frac{{{{(\ln x)}^2}}}{x},{\text{ }}x > 0$$.

The region $$R$$ is enclosed by the curve, the $$x$$-axis and the line $$x = e$$.

Let $${I_n} = \int_1^{\text{e}} {\frac{{{{(\ln x)}^n}}}{{{x^2}}}{\text{d}}x,{\text{ }}n \in \mathbb{N}}$$.

a.Given that the curve passes through the point $$(a,{\text{ }}0)$$, state the value of $$a$$.[1]

b.Use the substitution $$u = \ln x$$ to find the area of the region $$R$$.[5]
c.(i)     Find the value of $${I_0}$$.

(ii)     Prove that $${I_n} = \frac{1}{{\text{e}}} + n{I_{n – 1}},{\text{ }}n \in {\mathbb{Z}^ + }$$.

(iii)     Hence find the value of $${I_1}$$.[7]

d.Find the volume of the solid formed when the region $$R$$ is rotated through $$2\pi$$ about the $$x$$-axis.[5]

### Markscheme

$$a = 1$$    A1

[1 mark]

a.

$$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}$$    (A1)

$$\int {\frac{{{{(\ln x)}^2}}}{x}{\text{d}}x = \int {{u^2}{\text{d}}u} }$$    M1A1

area $$= \left[ {\frac{1}{3}{u^3}} \right]_0^1$$ or $$\left[ {\frac{1}{3}{{(\ln x)}^3}} \right]_1^{\text{e}}$$     A1

$$= \frac{1}{3}$$    A1

[5 marks]

b.

(i)     $${I_0} = \left[ { – \frac{1}{x}} \right]_1^{\text{e}}$$     (A1)

$$= 1 – \frac{1}{{\text{e}}}$$    A1

(ii)     use of integration by parts     M1

$${I_n} = \left[ { – \frac{1}{x}{{(\ln x)}^n}} \right]_1^{\text{e}} + \int_1^{\text{e}} {\frac{{n{{(\ln x)}^{n – 1}}}}{{{x^2}}}{\text{d}}x}$$    A1A1

$$= – \frac{1}{{\text{e}}} + n{I_{n – 1}}$$    AG

Note:     If the substitution $$u = \ln x$$ is used A1A1 can be awarded for $${I_n} = [ – {{\text{e}}^{ – u}}{u^n}]_0^1 + \int_0^1 n {{\text{e}}^{ – u}}{u^{n – 1}}{\text{d}}u$$.

(iii)     $${I_1} = – \frac{1}{{\text{e}}} + 1 \times {I_0}$$     (M1)

$$= 1 – \frac{2}{{\text{e}}}$$    A1

[7 marks]

c.

(d)     volume $$= \pi \int_1^{\text{e}} {\frac{{{{(\ln x)}^4}}}{{{x^2}}}{\text{d}}x{\text{ }}( = \pi {I_4})}$$     (A1)

EITHER

$${I_4} = – \frac{1}{{\text{e}}} + 4{I_3}$$    M1A1

$$= – \frac{1}{{\text{e}}} + 4\left( { – \frac{1}{{\text{e}}} + 3{I_2}} \right)$$    M1

$$= – \frac{5}{{\text{e}}} + 12{I_2} = – \frac{5}{{\text{e}}} + 12\left( { – \frac{1}{{\text{e}}} + 2{I_1}} \right)$$

OR

using parts $$\int_1^{\text{e}} {\frac{{{{(\ln x)}^4}}}{{{x^2}}}{\text{d}}x = – \frac{1}{{\text{e}}} + 4\int_1^{\text{e}} {\frac{{{{(\ln x)}^3}}}{{{x^2}}}{\text{d}}x} }$$     M1A1

$$= – \frac{1}{{\text{e}}} + 4\left( { – \frac{1}{{\text{e}}} + 3\int_1^{\text{e}} {\frac{{{{(\ln x)}^2}}}{{{x^2}}}{\text{d}}x} } \right)$$    M1

THEN

$$= – \frac{{17}}{{\text{e}}} + 24\left( {1 – \frac{2}{{\text{e}}}} \right) = 24 – \frac{{65}}{{\text{e}}}$$    A1

volume $$= \pi \left( {24 – \frac{{65}}{{\text{e}}}} \right)$$

[5 marks]

d.

### Question

a.Using the substitution $$x = \tan \theta$$ show that $$\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x = } \int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta }$$.[4]

b.Hence find the value of $$\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x}$$.[3]

### Markscheme

let $$x = \tan \theta$$

$$\Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}\theta }} = {\sec ^2}\theta$$     (A1)

$$\int {\frac{1}{{{{({x^2} + 1)}^2}}}{\text{d}}x = \int {\frac{{{{\sec }^2}\theta }}{{{{({{\tan }^2}\theta + 1)}^2}}}{\text{d}}\theta } }$$     M1

Note:     The method mark is for an attempt to substitute for both $$x$$ and $${\text{d}}x$$.

$$= \int {\frac{1}{{{{\sec }^2}\theta }}{\text{d}}\theta }$$ (or equivalent)     A1

when $$x = 0,{\text{ }}\theta = 0$$ and when $$x = 1,{\text{ }}\theta = \frac{\pi }{4}$$     M1

$$\int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta }$$    AG

[4 marks]

a.

$$\left( {\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x} = \int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } } \right) = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\left( {1 + \cos 2\theta } \right){\text{d}}\theta }$$   M1

$$= \frac{1}{2}\left[ {\theta + \frac{{\sin 2\theta }}{2}} \right]_0^{\frac{\pi }{4}}$$     A1

$$= \frac{\pi }{8} + \frac{1}{4}$$     A1

[3 marks]

b.

### Question

a.Use the substitution $$u = {x^{\frac{1}{2}}}$$ to find $$\int {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}}$$.[4]

b.Hence find the value of $$\frac{1}{2}\int\limits_1^9 {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}}$$, expressing your answer in the form arctan $$q$$, where $$q \in \mathbb{Q}$$.[3]

## Markscheme

a.$$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{2}{x^{ – \frac{1}{2}}}$$ (accept $${\text{d}}u = \frac{1}{2}{x^{ – \frac{1}{2}}}{\text{d}}x$$ or equivalent)       A1

substitution, leading to an integrand in terms of $$u$$     M1

$$\int {\frac{{2u{\text{d}}u}}{{{u^3} + u}}}$$ or equivalent      A1

= 2 arctan $$\left( {\sqrt x } \right)\left( { + c} \right)$$     A1

[4 marks]

b.

$$\frac{1}{2}\int\limits_1^9 {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}}$$ = arctan 3 − arctan 1     A1

tan(arctan 3 − arctan 1) = $$\frac{{3 – 1}}{{1 + 3 \times 1}}$$      (M1)

tan(arctan 3 − arctan 1) = $$\frac{1}{2}$$

arctan 3 − arctan 1 = arctan $$\frac{1}{2}$$     A1

[3 marks]

### Question

a.Find the value of the integral $$\int_0^{\sqrt 2 } {\sqrt {4 – {x^2}}f {\text{d}}x}$$ .[7]

b.

Find the value of the integral $$\int_0^{0.5} {\arcsin x {\text{d}}x}$$ .[5]

c.

Using the substitution $$t = \tan \theta$$ , find the value of the integral

$\int_0^{\frac{\pi }{4}} {\frac{{{\text{d}}\theta }}{{3{{\cos }^2}\theta + {{\sin }^2}\theta }}} {\text{ }}.$[7]

## Markscheme

a.let $$x = 2\sin \theta$$     M1

$${\text{d}}x = 2\cos \theta {\text{d}}\theta$$     A1

$$I = \int_0^{\frac{\pi }{4}} {2\cos \theta \times 2\cos \theta {\text{d}}\theta \,\,\,\,\,\left( { = 4\int_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } } \right)}$$     A1A1

Note: Award A1 for limits and A1 for expression.

$$= 2\int_0^{\frac{\pi }{4}} {(1 + \cos 2\theta ){\text{d}}\theta }$$     A1

$$= 2\left[ {\theta + \frac{1}{2}\sin 2\theta } \right]_0^{\frac{\pi }{4}}$$     A1

$$= 1 + \frac{\pi }{2}$$     A1

[7 marks]

b.

$$I = [x\arcsin x]_0^{0.5} – \int_0^{0.5} {x \times \frac{1}{{\sqrt {1 – {x^2}} }}{\text{d}}x}$$     M1A1A1

$$= [x\arcsin x]_0^{0.5} + \left[ {\sqrt {1 – {x^2}} } \right]_0^{0.5}$$     A1

$$= \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} – 1$$     A1

[5 marks]

c.

$${\text{d}}t = {\sec ^2}\theta {\text{d}}\theta {\text{ , }}\left[ {0,\frac{\pi }{4}} \right] \to [0,{\text{ 1]}}$$     A1(A1)

$$I = \int_0^1 {\frac{{\frac{{{\text{d}}t}}{{(1 + {t^2})}}}}{{\frac{3}{{(1 + {t^2})}} + \frac{{{t^2}}}{{(1 + {t^2})}}}}}$$     M1(A1)

$$= \int_0^1 {\frac{{{\text{d}}t}}{{3 + {t^2}}}}$$     A1

$$= \frac{1}{{\sqrt 3 }}\left[ {\arctan \left( {\frac{x}{{\sqrt 3 }}} \right)} \right]_0^1$$     A1

$$= \frac{\pi }{{6\sqrt 3 }}$$     A1

[7 marks]

### Question

Find
(a)  $$\int (2x+1)^{3}\: dx$$     (b)  $$\int (2x^{2}+1)^{3}\: dx$$     (c)  $$\int \frac{(2x^{2}+1)^{3}}{x}dx$$

Ans
(a)  $$\int (2x+1)^{3}dx=\frac{(2x+1)^{4}}{8}+c$$

(b)  $$\int (2x^{2}+1)^{3}\, dx=\int (8x^{6}+12x^{4}+6x^{2}+1)dx=\frac{8x^{7}}{7}+\frac{12x^{5}}{5}+2x^{3}+x+c$$

(c)  $$\int \frac{(2x^{2}+1)^{3}}{x}dx=\int (8x^{5}+12x^{3}+6x+\frac{1}{x})dx=\frac{8x^{6}}{3}+3x^{4}+3x^{2}+\textup{In}\left | x \right |+c$$

### Question

Use the double angle identities to calculate
(a)  $$\int \textup{cos}^{2}\; xdx$$     (b)  $$\int \textup{sin}^{2}\; xdx$$     (c)$$\int \textup{sin}\, x\, \textup{cos}\, xdx$$     (d)  $$\int \textup{sin}\, 5x\, \textup{cos}\, 5xdx$$

Ans
(a)  $$\int \textup{cos}^{2}\: xdx=\frac{1}{2}\int (1+\textup{cos2}x)dx=\frac{x}{2}+\frac{\textup{sin2}x}{4}+c$$

(b)  $$\int \textup{sin}^{2}\: xdx=\frac{1}{2}\int (1-\textup{cos2}x)dx=\frac{x}{2}-\frac{\textup{sin2}x}{4}+c$$

(c)  $$\int \textup{sin}\, x\, \textup{cos}\, xdx=\frac{1}{2}\int \textup{sin}\, 2xdx=-\frac{\textup{cos}2x}{4}+c$$

(d)  $$\int \textup{sin}\, 5x\, \textup{cos}\, 5xdx=\frac{1}{2}\int \textup{sin}10xdx=-\frac{\textup{cos}10x}{20}+c$$

### Questions

Find
(a)  $$\int \sqrt{x}(x^{3}-1)dx$$       (b)  $$\int \frac{3x^{2}-x+2\sqrt{x}}{3\sqrt{x}}dx$$

Ans
(a)  $$\int \sqrt{x}(x^{3}-1)dx=\int (x^{3.5}-x^{0.5})dx=\frac{x^{4.5}}{4.5}-\frac{x^{1.5}}{1.5}+c$$

(b)  $$\int \frac{3x^{2}-x+2\sqrt{2}}{3\sqrt{x}}dx=\int (x^{1.5}-\frac{1}{3}x^{0.5}+\frac{2}{3})dx=\frac{x^{2.5}}{2.5}-\frac{x^{1.5}}{4.5}+\frac{2}{3}x+c$$

### Question

Find    (a)   $$\int \frac{e^{3x}+2e^{x}+4}{3e^{2x}}dx$$          (b)  $$\int \frac{8^{x}+2^{x}+1}{4^{x}}dx$$.

Ans
(a)  $$\int \frac{e^{3x}+2e^{x}+4}{3e^{2x}}dx=\int (\frac{1}{3}e^{x}-\frac{2}{3}e^{-x}+\frac{4}{3}e^{-2x})dx=\frac{1}{3}e^{x}+\frac{2}{3}e^{-x}-\frac{2}{3}e^{-2x}+c$$

(b)  $$\int \frac{8^{x}+2^{x}+1}{4^{x}}dx=\int (2^{x}-2^{-x}+4^{-x})dx=\frac{2^{x}}{\textup{In}\, 2}+\frac{2^{-x}}{\textup{In}\, 2}-\frac{4^{-x}}{\textup{In}\, 4}+c$$

### Question

Let  $${f}'(x)=2e^{-x}+10\, \textup{sin}\, 5x+1$$.
Find $$f (x)$$ , given that the curve of this function passes through the point A(0,5).

Ans
$$f(x)=-2e^{-x}-2\textup{cos}5x+x+c$$
$$f(0)=5\Leftrightarrow -2-2+0+c=5\Leftrightarrow c=9$$
Hence,  $$f(x)=-2e^{-x}-2\textup{cos}5x+x+9$$

### Question

Let  $${f}”(x)=6ax+b$$.
Find $$f (x)$$ , given that the function has a maximum at A(1,5) and a point of inflection at B(0,3)

Ans
$${f}”(x)=6ax+b$$,     $${f}'(x)=3ax^{2}+bx+c$$     and      $$f(x)=ax^{3}+\frac{b}{2}x^{2}+cx+d$$

 Point of inflection at B(0,3) $$\Rightarrow {f}”(0)=0$$ and $$f(0)=3$$ $$\Rightarrow b=0$$ and $$d=3$$ Max at A(1,5) $$\Rightarrow {f}'(1)=0$$ and $$f(1)=5$$ $$\Rightarrow 3a+c=0$$ and $$a+c+3=5$$ $$\Rightarrow 3a+c=0$$ and $$a+c=2$$ $$\Rightarrow a=-1$$ and $$c=3$$

Therefore,  $$f(x)=-x^{3}+3x+3$$

### Question

Find
(a)  $$\int \frac{7}{16+x^{2}}dx$$          (b)  $$\int \frac{7}{1+16x^{2}}dx$$          (c)   $$\int \frac{7}{25+16x^{2}}dx$$         (d)  $$\int \frac{7}{2+x^{2}}dx$$         (e)  $$\int \frac{7}{2+3x^{2}}dx$$

Ans
(a)  $$\int \frac{7}{16+x^{2}}dx=\frac{7}{4}\textup{arctan}\frac{x}{4}+c$$

(b)  $$\int \frac{7}{1+16x^{2}}dx=\frac{7}{4}\textup{arctan}\, 4x+c$$

(c)  $$\int \frac{7}{25+16x^{2}}dx=\frac{7}{20}\textup{arctan}\frac{4x}{5}+c$$

(d)  $$\int \frac{7}{2+x^{2}}dx=\frac{7}{\sqrt{2}}\textup{arctan}\frac{x}{\sqrt{2}}+c$$

(e)  $$\int \frac{7}{2+3x^{2}}dx=\frac{7}{\sqrt{6}}\textup{arctan}\sqrt{\frac{3}{2}}x+c$$

### Question

Find
(a) $$\int \frac{7}{\sqrt{16-x^{2}}}dx$$         (b) $$\int \frac{7}{\sqrt{1-16x^{2}}}dx$$         (c) $$\int \frac{7}{\sqrt{25-16x^{2}}}dx$$         (d) $$\int \frac{7}{\sqrt{2-x^{2}}}dx$$         (e) $$\int \frac{7}{\sqrt{2-3x^{2}}}dx$$

Ans
(a) $$\int \frac{7}{\sqrt{16-x^{2}}}dx=7\, \textup{arcsin}\frac{x}{4}+c$$

(b) $$\int \frac{7}{\sqrt{1-16x^{2}}}dx=\frac{7}{4}\, \textup{arcsin}\, 4x+c$$

(c) $$\int \frac{7}{\sqrt{25-16x^{2}}}dx=\frac{7}{4}\, \textup{arcsin}\frac{4x}{5}+c$$

(d) $$\int \frac{7}{\sqrt{2-x^{2}}}dx=7\, \textup{arcsin}\frac{x}{\sqrt{2}}+c$$

(e) $$\int \frac{7}{\sqrt{2-3x^{2}}}dx=\frac{7}{\sqrt{3}}\, \textup{arcsin}\sqrt{\frac{3}{2}}x+c$$

### Question

Find        (a) $$\int \frac{7}{(x-1)^{2}+4}dx$$              (b) $$\int \frac{7}{4(x-1)^{2}+1}dx$$

Ans
(a) $$\int \frac{7}{(x-1)^{2}+4}dx=\frac{7}{2}\textup{arctan}\frac{x-1}{2}+c$$            (b) $$\int \frac{7}{4(x-1)^{2}+1}dx=\frac{7}{2}\textup{arctan}(2x-2)+c$$

### Question

Find  (a)      $$\int \frac{7}{\sqrt{4-(x-1)^{2}}}dx$$              (b) $$\int \frac{7}{\sqrt{1-4(x-1)^{2}}}dx$$

Ans
(a)   $$\int \frac{7}{\sqrt{4-(x-1)^{2}}}dx=7\textup{arcsin}\frac{x-1}{2}+c$$             (b)   $$\int \frac{7}{\sqrt{1-4(x-1)^{2}}}dx=\frac{7}{2}\textup{arcsin}(2x-2)+c$$

### Question

(a) Express  $$\frac{2}{x^{2}-10x+24}$$ in partial fractions
(b) Express  $$x^{2}-10x+26$$ in vertex form
(c) Find the integrals
(i) $$\int \frac{2}{x^{2}-10x+24}dx$$             (ii) $$\int \frac{2}{x^{2}-10x+25}dx$$             (iii) $$\int \frac{2}{x^{2}-10x+26}dx$$

Ans
(a) $$\frac{2}{x^{2}-10x+24}=\frac{A}{x-6}+\frac{B}{x-4}\Rightarrow 2=A(x-4)+B(x-6)$$
For $$x=6,\, 2A=2\Leftrightarrow A=1$$
For $$x=4,\, -2B=2\Leftrightarrow B=-1$$
Hence, $$\frac{2}{x^{2}-10x+24}=\frac{1}{x-6}-\frac{1}{x-4}$$

(b) $$x^{2}-10x+26=(x-5)^{2}+1$$

(c) (i) $$\int \frac{2}{x^{2}-10x+24}dx=\int \frac{1}{x-6}-\frac{1}{x-4}dx=\textup{In}\left | x-6 \right |-\textup{In}\left | x-4 \right |+c=\, \textup{In}\left | \frac{x-6}{x-4} \right |+c$$
(ii) $$\int \frac{2}{x^{2}-10x+25}dx=\int \frac{2}{(x-5)^{2}}dx=-\frac{2}{x-5}+c$$
(iii) $$\int \frac{2}{x^{2}-10x+26}dx=\int \frac{2}{(x-5)^{2}+1}dx=2\, \textup{arctan}(x-5)+c$$

### Question

Let $$f(x)=\frac{x^{2}+3x+12}{x(x+2)^{2}}$$

(a) Show that $$f(x)=\frac{3}{x}-\frac{2}{(x+2)}-\frac{5}{(x+2)^{2}}$$
(b) Hence find $$\int f(x)dx$$

Ans
(b) $$\int f(x)dx=3\, \textup{In}\left | x \right |-2\, \textup{In}\left | x+2 \right |+\frac{5}{x+2}+c$$

### Question

Complete the following table.   [Look at the formula booklet!]

 Integral Result $$\int \textup{sec}^{2}\, xdx$$ $$\int (\textup{tan}^{2}\, x+1)dx$$ $$\int \frac{5}{\textup{cos}^{2}\, x}dx$$ $$\int \textup{csc}^{2}\, xdx$$ $$\int (\textup{cot}^{2}\, x+1)dx$$ $$\int \frac{5}{\textup{sin}^{2}\, x}dx$$ $$\int \textup{sec}\, x\, \textup{tan}\, xdx$$ $$\int \textup{csc}\, x\, \textup{cot}\, xdx$$ $$\int \textup{tan}^{2}\, xdx$$ $$\int \textup{cot}^{2}\, xdx$$

Ans

 Integral Result $$\int \textup{sec}^{2}\, xdx$$ $$\textup{tan}\, x+c$$ $$\int (\textup{tan}^{2}\, x+1)dx$$ $$\textup{tan}\, x+c$$ $$\int \frac{5}{\textup{cos}^{2}\, x}dx$$ $$5\, \textup{tan}\, x+c$$ $$\int \textup{csc}^{2}\, xdx$$ $$-\textup{cot}\, x+c$$ $$\int (\textup{cot}^{2}\, x+1)dx$$ $$-\textup{cot}\, x+c$$
 Integral Result $$\int \frac{5}{\textup{sin}^{2}\, x}dx$$ $$-5\, \textup{cot}\, x+c$$ $$\int \textup{sec}\, x\, \textup{tan}\, xdx$$ $$\textup{sec}\, x+c$$ $$\int \textup{csc}\, x\, \textup{cot}\, xdx$$ $$-\, \textup{csc}\, x+c$$ $$\int \textup{tan}^{2}\, xdx$$ $$\textup{tan}\, x\, -x+c$$ $$\int \textup{cot}^{2}\, xdx$$ $$-\textup{cot}\, x\, -x+c$$

### Question

Let $$f(t)=t^{\frac{1}{3}}\left ( 1-\frac{1}{2t^{\frac{5}{3}}} \right )$$.  Find $$\int f(t)\, \textup{d}t$$.                                                      (Total 3 marks)

Ans
$$\int t^{\frac{1}{3}}\left ( 1-\frac{1}{2t^{\frac{2}{5}}} \right )\textup{d}t=\int t^{\frac{1}{3}}\left ( 1-\frac{t^{\frac{5}{3}}}{2} \right )\textup{d}t=\int \left ( t^{\frac{1}{3}}-\frac{t^{\frac{4}{3}}}{2} \right )\textup{d}t$$                   (M1)
$$=\frac{3}{4}t^{\frac{4}{3}}+\frac{3}{2}t^{\frac{1}{3}}+c$$                                                                                                                    (M1)(A1)  (C3)
Note: Do not penalise for the absence of +C.                                                                                          [3]

### Question

The function $${f}’$$ is given by $${f}'(x)=2\, \textup{sin}\left ( 5x-\frac{\pi }{2} \right )$$.
(a) Write down $${f}”(x)$$.              (b)  Given that $$f\left ( \frac{\pi}{2} \right )=1$$, find $$f(x)$$.
(Total 6 marks)

Ans
(a)  Using the chain rule $${f}'(x)=\left ( 2\, \textup{cos}\left ( 5x-\frac{\pi }{2} \right ) \right )5$$                                  (M1)
$$=10\, \textup{cos}\left ( 5x-\frac{\pi }{2} \right )$$                                                                                                             A1     2

(b)  $$f(x)=\int {f}'(x)dx=-\frac{2}{5}\textup{cos}\left ( 5x-\frac{\pi }{2} \right )+c$$                                            A1
Substituting to find $$c,f\left ( \frac{\pi }{2} \right )=\frac{2}{5}\textup{cos}\left ( 5\left ( \frac{\pi }{2} \right )-\frac{\pi }{2} \right )+c=1$$                 M1
$$c=1+\frac{2}{5}\textup{cos}2\pi =1+\frac{2}{5}=\frac{7}{5}$$                                                                               (A1)
$$f(x)=\frac{2}{5}\textup{cos}\left ( 5x-\frac{\pi }{2} \right )+\frac{7}{5}$$                                                                                      A1      4                                          [6]

### Question

Find the indefinite integrals
(a)  $$\int \frac{1}{x^{2}+2}dx$$           (b)  $$\int \frac{x}{x^{2}+2}dx$$           (c)  $$\int \frac{x^{2}}{x^{2}+2}dx$$           (d)  $$\int \frac{x^{3}}{x^{2}+2}dx$$

Ans
(a) $$\frac{1}{\sqrt{2}}\textup{arctan}\frac{x}{\sqrt{2}}+c$$     (b) $$\frac{1}{2}\textup{In}(x^{2}+2)+c$$     (c) $$x-\frac{2}{\sqrt{2}}\textup{arctan}\frac{x}{\sqrt{2}}+c$$     (d) $$\frac{x^{2}}{2}-\textup{In}(x^{2}+2)+c$$

### Question

Find the indefinite integral  $$\int \frac{12x+10}{3x^{2}+5x+1}dx$$

Ans
$$2\, \textup{In}(3x^{2}+5x+1)+c$$

### Question

Find the indefinite integrals  (a) $$\int \textup{tan}\, xdx$$                  (b) $$\int \textup{cot}\, xdx$$

Ans
(a) $$\int \textup{tan}\, xdx=\int \frac{\textup{sin}\, x}{\textup{cos}\, x}dx=-\textup{In}\left | \textup{cos}\, x \right |+c$$    (use the substitution $$u=\textup{cos}\, x$$)
(b) $$\int \textup{cot}\, xdx=\int \frac{\textup{cos}\, x}{\textup{sin}\, x}dx=\textup{In}\left | \textup{sin}\, x \right |+c$$    (use the substitution $$u=\textup{sin}\, x$$)

### Question

Find the following indefinite integrals by using the substitution  $$u=x^{2}+5$$
(a) $$\int \frac{x^{5}}{x^{2}+5}dx$$            (b) $$\int \frac{x^{3}}{\sqrt{x^{2}+5}}dx$$          (c) $$\int x^{3}(x^{2}+5)^{5}dx$$

Ans
(a) $$\frac{25}{2}\textup{In}(x^{2}+5)+\frac{(x^{2}+5)^{2}}{4}-5x^{2}+c$$       (b) $$\frac{(x^{2}-10)\sqrt{x^{2}+5}}{3}+c$$       (c) $$\frac{(x^{2}+5)^{6}(6x^{2}-5)}{84}+c$$

### Question

Find the indefinite integrals
(a) $$\int \frac{x}{x^{2}+9}dx$$           (b) $$\int \frac{x}{(x^{2}+9)^{2}}dx$$           (c) $$\int \frac{x^{2}+x+12}{x^{2}+9}dx$$

Ans(a) $$\frac{1}{2}\textup{In}(x^{2}+9)+c$$           (b) $$-\frac{1}{2(x^{2}+9)}+c$$           (c) $$\frac{1}{2}\textup{In}(x^{2}+9)+x+\textup{arctan}\frac{x}{3}+c$$

### Question

Find the indefinite integrals
(a) $$\int \textup{tan}^{5}\, x\, \textup{sec}^{2}\, xdx$$          (b) $$\int \frac{\textup{sec}^{2}\, x}{\sqrt{\textup{tan}\, x}}dx$$         (c) $$\int \frac{\textup{sec}^{2}\, x}{e^{\textup{tan}\, x}}dx$$         (d) $$\int \frac{\textup{sin}^{5}\, x}{\textup{cos}^{7}\, x}dx$$

Ans

(a) $$\frac{\textup{tan}^{6}\, x}{6}+c$$        (b) $$2\sqrt{\textup{tan}\, x}+c$$       (c) $$-\frac{1}{e^{\textup{tan}\, x}}+c$$       (d) $$\frac{\textup{tan}^{6}\, x}{6}+c$$

### Question

(a) Calculate $$\int \frac{1}{5+x^{2}}dx$$ , by using the substitution $$x=\sqrt{5}\, \textup{tan}\, \theta$$
(b) Calculate  (i) $$\int \frac{1}{\sqrt{5-x^{2}}}dx$$ ,   (ii) $$\int \sqrt{5-x^{2}\, dx}$$   by using the substitution $$x=\sqrt{5}\, \textup{sin}\, \theta$$

Ans
(a) $$\frac{1}{\sqrt{5}}\textup{arctan}\frac{x}{\sqrt{5}}+c$$         (b)  (i) $$\textup{arcsin}\frac{x}{\sqrt{5}}+c$$ ,      (ii) $$\frac{5}{2}\textup{arcsin}\frac{x}{\sqrt{5}}+\frac{x\sqrt{5-x^{2}}}{2}+c$$

### Question

Calculate $$\int \sqrt{\frac{x}{4-x}}dx$$ , by using the substitution $$x=4\, \textup{sin}^{2}\, \theta$$
Express your answer in the form $$A\, \textup{arcsin}\frac{\sqrt{x}}{2}+B\sqrt{4x-x^{2}}+c$$

Ans
$$4\, \textup{arcsin}\frac{\sqrt{x}}{2}-\sqrt{4x-x^{2}}+c$$

### Question

(a) Show that $$\frac{2x+4}{(x^{2}+4)(x-2)}=\frac{1}{x-2}-\frac{x}{x^{2}+4}$$
(b) Hence find $$\int \frac{2x+4}{(x^{2}+4)(x-2)}dx$$

Ans
(b) $$\textup{In}(x-2)-\frac{1}{2}\textup{In}(x^{2}+4)+c$$

### Question

Using the substitution $$y=2-x$$, or otherwise, find $$\int \left ( \frac{x}{2-x} \right )^{2}dx$$.                                               (Total 6 marks)

Ans
$$I=\int \left ( \frac{2-y}{y} \right )^{2}(-dy)$$                                                                                          (M1)(A1)
$$=\int \left ( \frac{4}{y^{2}}-\frac{4}{y}+1 \right )\textup{d}y=\frac{4}{y}+4\, \textup{In}\left | y \right |-y+c$$                               (A1)(A1)(A1)
$$=\frac{4}{2-x}+4\textup{In}\left | 2-x \right |-(2-x)+c$$                                                       (A1) (C6)
Note: c and modulus signs not required.                                                                                                               [6]

### Question

Using the substitution  $$u=\frac{1}{2}x+1$$,or otherwise, find the integral $$\int x\sqrt{\frac{1}{2}x+1}\: dx$$.                                               (Total 4 marks)

Ans
Let $$u=\frac{1}{2}x+1\Leftrightarrow x=2(u-1)\Rightarrow \frac{\textup{d}x}{\textup{d}u}=2$$
Then $$\int x\left ( \frac{1}{2}x+1 \right )^{1/2}dx$$
$$=\int 2(u-1)\times u^{1/2}\times 2\textup{d}u$$                                                        (M1)
$$=4\int (u^{3/2}-u^{1/2})\textup{d}u$$                                                                      (A1)
$$=4\left [ \frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2} \right ]+C$$                                                          (M1)
$$=4\left [ \frac{2}{5}\left ( \frac{1}{2}x+1 \right )^{5/2}-\frac{2}{3}\left ( \frac{1}{2}x+1 \right )^{3/2} \right ]+C$$                 (A1)                                 [4]

### Question

Use the substitution $$u=x+2$$ to find $$\int \frac{x^{3}}{(x+2)^{2}}\textup{d}x$$.                                           (Total 6 marks)

Ans
Substituting $$u=x+2\Rightarrow u-2=x,\textup{d}u=\textup{d}x$$                      (M1)
$$\int \frac{x^{3}}{(x+2)^{2}}\textup{d}x=\frac{(u-2)^{3}}{u^{2}}\textup{d}u$$                                                                             A1
$$=\int \frac{u^{3}-6u^{2}+12u-8}{u^{2}}\textup{d}u$$                                                                                    A1
$$\int u\, \textup{d}u+\int (-6)\textup{d}u+\int \frac{12}{u}\textup{d}u+\int 8u^{-2}\textup{d}u$$                                A1
$$=\frac{u^{2}}{2}-6u+12\textup{In}\left | u \right |+8u^{-1}+c$$                                                   A1
$$=\frac{(x+2)^{2}}{2}-6(x+2)+12\, \textup{In}\left | x+2 \right |+\frac{8}{x+2}+c$$                  A1                                 [6]

### Question

Find $$\int \frac{e^{x}}{e^{2x}+9}dx$$.                                                     (Total 5 marks)
Extra question
Find in a similar way the integral $$\int \frac{3^{x}}{9^{x}+9}dx$$

Ans
$$\textup{Let }u=e^{x}$$                                                                                                                                       M1
$$\textup{d}u=e^{x}\textup{d}x\, (\textup{or equivalent})$$                                                                                                A1
$$\textup{When}\, x=0,u=1\, \textbf{and}\, \textup{when}\, x=\textup{In}\, 3, u=3$$                                                 (A1)
$$\int_{0}^{\textup{In}3}\frac{e^{x}}{e^{2x}+9}dx=\int_{1}^{3}\frac{1}{u^{2}+9}\textup{d}u$$                                                                                                 A1
$$=\frac{1}{3}\left [ \textup{arctan}\frac{u}{3} \right ]^{3}_{1}$$                                                                                                                          A1
$$=\frac{1}{3}\left ( \textup{arctan}1-\textup{arctan}\frac{1}{3} \right )\left ( =\frac{\pi }{12}-\frac{1}{3}\textup{arctan}\frac{1}{3},\frac{1}{3}\textup{arctan}\frac{1}{2} \right )$$                 A1              N0                  [6]
Extra question
Find in a similar way the integral $$\int \frac{3^{x}}{9^{x}+9}dx$$

### Question

By using an appropriate substitution find $$\int \frac{\textup{tan}(\textup{In}\, y)}{y}dy,y> 0$$.                   (Total 6 marks)
Extra question
The integral $$\int \frac{\textup{cot}(\textup{In}\, y)}{y}dy,y> 0$$ can be found in a similar way. Write down the result!

Ans
$$\textup{Let}\, u=\textup{In}\, y\Rightarrow \textup{d}u=\frac{1}{y}\textup{d}y$$                                 A1(A1)
$$\int \frac{\textup{tan}(\textup{In}\, y)}{y}\textup{d}y=\int \textup{tan}\, u\, \textup{d}u$$                                   A1
$$\int \frac{\textup{sin}\, u}{\textup{cos}\, u}\textup{d}u=-\textup{In}\left | \textup{cos}\, u \right |+c$$                              A1
EITHER
$$\int \frac{\textup{tan}(\textup{In}\, y)}{y}\textup{d}y=-\textup{In}\left | \textup{cos}(\textup{In}\, y) \right |+c$$             A1A1
OR
$$\int \frac{\textup{tan}(\textup{In}\, y)}{y}\textup{d}y=\textup{In}\left | \textup{sec}(\textup{In}\, y) \right |+c$$                 A1A1                                                                         [6]Extra question
$$\textup{In}\left | \textup{sin}(\textup{In}\, y) \right |+c$$

### Question

Using the substitution $$2x=\textup{sin}\theta$$, or otherwise, find $$\int \left ( \sqrt{1-4x^{2}} \right )dx$$.                                                  (Total 6 marks)

$$\int \left (\sqrt{1-4x^{2}} \right )\textup{d}x$$
$$\textup{Let}\, 2x=\textup{sin}\theta \Rightarrow 2\frac{\textup{d}x}{\textup{d}\theta }=\textup{cos}\, \theta \Rightarrow \textup{d}x=\frac{1}{2}\textup{cos}\, \theta \, \textup{d}\, \theta$$
$$\Rightarrow \int \left ( \sqrt{1-4x^{2}} \right )\textup{d}x=\int \sqrt{1-\textup{sin}^{2}\theta \frac{1}{2}}\textup{cos}\,\theta \textup{d}\, \theta$$
$$=\int \frac{1}{2}\textup{cos}^{2}\theta \, \textup{d}\, \theta$$                                                                                      (A1)
$$\int \frac{1}{4}(\textup{cos}\, 2\theta +1)\textup{d}\theta$$                                                                              (A1)
$$=\frac{1}{8}\textup{sin}\, 2\theta +\frac{\theta }{4}+C$$                                                                          (A1)(A1)
$$=\frac{1}{4}\left [ 2x\sqrt{1-4x^{2}}+\textup{arcsin}\, 2x \right ]+C$$                                  (A1)(A1)    (C6)                                        [6]