IBDP Maths analysis and approaches Topic: SL 5.10 Indefinite integral of xn , sinx, cosx, and ex HL Paper 1

Question

The acceleration, a ms-2 , of a particle moving in a horizontal line at time t seconds, t ≥ 0 , is given by a = – (1+v) where v ms-1 is the particle’s velocity and v > -1.
At t = 0 , the particle is at a fixed origin O and has initial velocity v0 ms-1 .

(a) By solving an appropriate differential equation, show that the particle’s velocity at time t is given by v ( t ) = (1 + v0) e-t – 1 . [6]

(b) Initially at O, the particle moves in the positive direction until it reaches its maximum displacement from O. The particle then returns to O.

Let s metres represent the particle’s displacement from O and smax its maximum displacement from O.
(i) Show that the time T taken for the particle to reach smax satisfies the equation eT = 1 + v0 .
(ii) By solving an appropriate differential equation and using the result from part (b) (i), find an expression for smax in terms of v0 . [7]

Let v (T – k) represent the particle’s velocity k seconds before it reaches smax , where v (T – k) = (1 + v0) e-(T – k) – 1 .

(c) By using the result to part (b) (i), show that v (T – k) = ek – 1 . [2]

Similarly, let v (T + k) represent the particle’s velocity k seconds after it reaches smax .

(d) Deduce a similar expression for v (T + k) in terms of k . [2]

(e) Hence, show that v (T – k) + v (T + k) ≥ 0 . [3]

Ans:

(a) Since $a=\frac{\text{d}v}{\text{d}t}$, and $a=-\left(1+v\right)$, we have
$$\begin{eqnarray} \frac{\text{d}v}{\text{d}t} = -1-v \nonumber \\ \frac{\text{d}v}{\text{d}t}+v = -1. \end{eqnarray}$$
Let $\text{I}\left(x\right)=\text{e}^{\int 1 \text{d}t}=\text{e}^t$.<br>
Then, we have
$$\begin{eqnarray} \frac{\text{d}}{\text{d}t}\left(v\text{e}^t\right) = -\text{e}^t \nonumber \\ v\text{e}^t = -\int\text{e}^t \text{d}t \nonumber \\ v\text{e}^t = -\text{e}^t+c \nonumber \\ v = c\text{e}^{-t}-1. \end{eqnarray}$$
Since $v=v_0$ when $t=0$, we have $c=1+v_0$, i.e., $v=\left(1+v_0\right)\text{e}^{-t}-1$.<br>
(b)(i) At $s_{\text{max}}$, $t=T$, i.e., we have
$$\begin{eqnarray} v\left(T\right) = 0 \nonumber \\ \left(1+v_0\right)\text{e}^{-T}-1 = 0 \nonumber \\ \left(1+v_0\right)\text{e}^{-T} = 1 \nonumber \\ 1+v_0 = \text{e}^T. \end{eqnarray}$$
(b)(ii) Since $\frac{\text{d}s}{\text{d}t}=\left(1+v_0\right)\text{e}^{-t}-1$, integrating both sides with respect to $t$, we have
$$\begin{eqnarray} s &=& \int\left(1+v_0\right)\text{e}^{-t}-1\text{d}t \nonumber \\ &=& -\left(1+v_0\right)\text{e}^{-t}-t+c. \end{eqnarray}$$
When $t=0$, $s=0$, i.e.,
$$\begin{eqnarray} 0 = -\left(1+v_0\right)+c \nonumber \\ c = \left(1+v_0\right). \end{eqnarray}$$
Thus, $s=\left(1+v_0\right)-\left(1+v_0\right)\text{e}^{-t}-t$.<br>
Since from (b)(i) we have $1+v_0=\text{e}^T$, $T=\ln\left(1+v_0\right)$, i.e., $s_{\text{max}}=v_0-\ln \left(1+v_0\right)$.<br>
(c) Since $\text{e}^T=1+v_0$, we have $\left(1+v_0\right)\text{e}^{-T}=1$, i.e.,
$$\begin{eqnarray} v\left(T-k\right) &=& \left(1+v_0\right)\text{e}^{-\left(T-k\right)}-1 \nonumber \\ &=& \left(1+v_0\right)\text{e}^{-T}\text{e}^k-1 \nonumber \\ &=& \text{e}^k-1. \end{eqnarray}$$
(d) Similarly, as in (c),
$$\begin{eqnarray} v\left(T+k\right) &=& \left(1+v_0\right)\text{e}^{-\left(T+k\right)}-1 \nonumber \\ &=& \left(1+v_0\right)\text{e}^{-T}\text{e}^{-k}-1 \nonumber \\ &=& \text{e}^{-k}-1. \end{eqnarray}$$
(e)
$$\begin{eqnarray} v\left(T-k\right)+v\left(T+k\right) &=& \text{e}^k-1+\text{e}^{-k}-1 \nonumber \\ &=& \text{e}^k+\text{e}^{-k}-2 \nonumber \\ &=& \frac{\text{e}^{2k}-2\text{e}^k+1}{\text{e}^k} \nonumber \\ &=& \frac{\left(\text{e}^k-1\right)^2}{\text{e}^k} \geq 0. \end{eqnarray}$$

Question

Find $$\int {(1 + {{\tan }^2}x){\text{d}}x}$$.[2]

a.

Find $$\int {{{\sin }^2}x{\text{d}}x}$$.[3]

b.

Markscheme

$$\int {(1 + {{\tan }^2}x){\text{d}}x} = \int {{{\sec }^2}x{\text{d}}x = \tan x( + c)}$$     M1A1

[2 marks]

a.

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

$$= \frac{x}{2} – \frac{{\sin 2x}}{4}( + c)$$     A1

Note:     Allow integration by parts followed by trig identity.

Award M1 for parts, A1 for trig identity, A1 final answer.

[3 marks]

Total [5 marks]

b.

Question

A function $$f$$ is defined by $$f(x) = \frac{{3x – 2}}{{2x – 1}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne \frac{1}{2}$$.

Find an expression for $${f^{ – 1}}(x)$$.[4]

a.

Given that $$f(x)$$ can be written in the form $$f(x) = A + \frac{B}{{2x – 1}}$$, find the values of the constants $$A$$ and $$B$$.[2]

b.

Hence, write down $$\int {\frac{{3x – 2}}{{2x – 1}}} {\text{d}}x$$.[1]

c.

Markscheme

$$f:x \to y = \frac{{3x – 2}}{{2x – 1}}\;\;\;{f^{ – 1}}:y \to x$$

$$y = \frac{{3x – 2}}{{2x – 1}} \Rightarrow 3x – 2 = 2xy – y$$     M1

$$\Rightarrow 3x – 2xy = – y + 2$$     M1

$$x(3 – 2y) = 2 – y$$

$$x = \frac{{2 – y}}{{3 – 2y}}$$     A1

$$\left( {{f^{ – 1}}(y) = \frac{{2 – y}}{{3 – 2y}}} \right)$$

$${f^{ – 1}}(x) = \frac{{2 – x}}{{3 – 2x}}\;\;\;\left( {x \ne \frac{3}{2}} \right)$$     A1

Note:     $$x$$ and $$y$$ might be interchanged earlier.

Note:     First M1 is for interchange of variables second M1 for manipulation

Note:     Final answer must be a function of $$x$$

[4 marks]

a.

$$\frac{{3x – 2}}{{2x – 1}} = A + \frac{B}{{2x – 1}} \Rightarrow 3x – 2 = A(2x – 1) + B$$

equating coefficients $$3 = 2A$$ and $$– 2 = – A + B$$     (M1)

$$A = \frac{3}{2}$$ and $$B = – \frac{1}{2}$$     A1

Note:     Could also be done by division or substitution of values.

[2 marks]

b.

$$\int {f(x){\text{d}}x = \frac{3}{2}x – \frac{1}{4}\ln \left| {2x – 1} \right| + c}$$     A1

Note:     accept equivalent e.g. $$\ln \left| {4x – 2} \right|$$

[1 mark]

Total [7 marks]

c.

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)  $$\int \frac{7}{2+3x^{2}}dx$$

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

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

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

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

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]