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

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▶️Answer/Explanation

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 $

▶️Answer/Explanation

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

▶️Answer/Explanation

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 $\int \frac{2x+4}{(x^{2}+4)(x-2)}dx$

▶️Answer/Explanation

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)

▶️Answer/Explanation

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)

▶️Answer/Explanation

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)

▶️Answer/Explanation

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 $\int \frac{3^{x}}{9^{x}+9}dx$

▶️Answer/Explanation

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!

▶️Answer/Explanation

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)

▶️Answer/Explanation

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) = e^x 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) = e^x 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 $\lim_{x\rightarrow 0}\frac{e^{x}cos x – 1 – x}{x^{3}}.$

▶️Answer/Explanation

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 -2e^x sin x .

Note: Accept working with each side separately to obtain -4e^x 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]

▶️Answer/Explanation

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 $

▶️Answer/Explanation

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 $

▶️Answer/Explanation

Ans:

Question

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

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

▶️Answer/Explanation

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

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

$\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]

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

▶️Answer/Explanation

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]

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

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]

▶️Answer/Explanation

Markscheme

(i)

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]

$\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]

(i)

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]

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]

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

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]

▶️Answer/Explanation

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

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

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

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

$ \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]

$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]

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]

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]

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

Find the range of f.[2]

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]

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

▶️Answer/Explanation

Markscheme

$4{(x – 0.5)^2} + 4$ A1A1

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

[2 marks]

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]

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]

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

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

[2 marks]

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]

$\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]

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)} $.

▶️Answer/Explanation

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} $.

▶️Answer/Explanation

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} $.

▶️Answer/Explanation

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

${{\text{e}}^x} = u – 3$

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.

▶️Answer/Explanation

Markscheme

EITHER

$x = \arctan t$ (M1)

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

$t = \tan x$

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

▶️Answer/Explanation

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]

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

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

Find the range of $f$.[3]

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

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

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]

▶️Answer/Explanation

Markscheme

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

$ = – x\sqrt {1 – {x^2}} $

$ = – f(x)$ R1

hence $f$ is odd AG

[2 marks]

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

[3 marks]

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

$y$-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]

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]

${\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]

$\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]

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]

▶️Answer/Explanation

Markscheme

$a = 1$ A1

[1 mark]

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

(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]

(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)$

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]

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]

▶️Answer/Explanation

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]

$\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]

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]

▶️Answer/Explanation

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]

$\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]

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

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]

▶️Answer/Explanation

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]

$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]

${\text{d}}t = {\sec ^2}\theta {\text{d}}\theta {\text{ , }}\left[ {0,\frac{\pi }{4}} \right] \to [0,{\text{ 1]}}$ 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) $\int \frac{(2x^{2}+1)^{3}}{x}dx$

▶️Answer/Explanation

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) $\int \textup{sin}\, 5x\, \textup{cos}\, 5xdx$

▶️Answer/Explanation

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

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

▶️Answer/Explanation

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) $\int \frac{8^{x}+2^{x}+1}{4^{x}}dx$.

▶️Answer/Explanation

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).

▶️Answer/Explanation

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)

▶️Answer/Explanation

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) $\int \frac{7}{2+3x^{2}}dx$

▶️Answer/Explanation

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

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

(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) $\int \frac{7}{\sqrt{2-3x^{2}}}dx$

▶️Answer/Explanation

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

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

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

▶️Answer/Explanation

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

▶️Answer/Explanation

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) $\int \frac{2}{x^{2}-10x+26}dx$

▶️Answer/Explanation

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 $\int f(x)dx$

▶️Answer/Explanation

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

▶️Answer/Explanation

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)

▶️Answer/Explanation

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)

▶️Answer/Explanation

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) $\int \frac{x^{3}}{x^{2}+2}dx$

▶️Answer/Explanation

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

▶️Answer/Explanation

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

Question

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

▶️Answer/Explanation

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) $\int x^{3}(x^{2}+5)^{5}dx$

▶️Answer/Explanation

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) $\int \frac{x^{2}+x+12}{x^{2}+9}dx$

▶️Answer/Explanation

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) $\int \frac{\textup{sin}^{5}\, x}{\textup{cos}^{7}\, x}dx$

▶️Answer/Explanation

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

▶️Answer/Explanation

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

▶️Answer/Explanation

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 $\int \frac{2x+4}{(x^{2}+4)(x-2)}dx$

▶️Answer/Explanation

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)

▶️Answer/Explanation

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)

▶️Answer/Explanation

Question

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

▶️Answer/Explanation

Question

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

▶️Answer/Explanation

Question

▶️Answer/Explanation

Question

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

▶️Answer/Explanation

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	\(c=3\)

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

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

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