Answer the question [ 4 mark question ]

Question 14 Marks

Integrate the function: $\frac{\sqrt{\tan x}}{\sin x \cos x}$

Answer

Let $I=\int \frac{\sqrt{\tan x}}{\sin x \cos x}$
$=\int \frac{\sqrt{\tan x} \cdot \cos x}{\sin x \cos x \cdot \cos x} d x$
$= \int \frac{\sqrt{\tan x}}{\tan x \cos ^{2} x} d x$
$= \int \frac{\sec ^{2} x d x}{\sqrt{\tan x}}$
Let tan x = t $\Rightarrow$ sec²x dx = dt
$\Rightarrow I=\int \frac{d t}{\sqrt{t}}$
$=2 \sqrt{t}+C$
$=2 \sqrt{\tan x}+c$

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Question 24 Marks

Integrate the function: $\frac{1}{{1 - \tan x}}$

Answer

Let $I = \int {\frac{1}{{1 - \tan x}}} dx$

$= \int {\frac{1}{{1 - \frac{{\sin x}}{{\cos x}}}}} dx$

$= \int {\frac{1}{{\left( {\frac{{\cos x - \sin x}}{{\cos x}}} \right)}}} dx$

$= \int {\frac{{\cos x}}{{\cos x - \sin x}}} dx$

$= \frac{1}{2}\int {\frac{{2\cos x}}{{\cos x - \sin x}}} dx$

$= \frac{1}{2}\int {\frac{{\cos x + \cos x}}{{\cos x - \sin x}}} dx$

Adding and subtracting sin x in the numerator,

$= \frac{1}{2}\int {\frac{{\cos x - \sin x + \sin x + \cos x}}{{\cos x - \sin x}}} dx$

$= \frac{1}{2}\int {\frac{{\left( {\cos x - \sin x} \right) + \left( {\sin x + \cos x} \right)}}{{\cos x - \sin x}}} dx$

$= \frac{1}{2}\int {\frac{{\cos x - \sin x}}{{\cos x - \sin x}}} + \frac{{\sin x + \cos x}}{{\cos x - \sin x}}dx$

$= \frac{1}{2}\int {\left( {1 + \frac{{\sin x + \cos x}}{{\cos x - \sin x}}} \right)} dx$

$= \frac{1}{2}\left[ {\int {1dx-\int {\frac{{ - \sin x - \cos x}}{{\cos x - \sin x}}} } } \right]dx$

$= \frac{1}{2}\left[ {x - \log \left| {\cos x - \sin x} \right|} \right] + c$ $[\because \int \frac{f'(x)}{f(x)} dx=log|f(x)|+c]$

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Question 34 Marks

Integrate the function: $\frac{1}{1+\cot x}$

Answer

Let $I=\int \frac{1}{1+\cot x} d x$
$=\int \frac{1}{1+\frac{\cos x}{\sin x}} d x$
$=\int \frac{\sin x}{\sin x+\cos x} d x$
$= \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{\sin x+\cos x} d x$
$=\frac{1}{2} \int 1 . d x+\frac{1}{2} \int \frac{(\sin x-\cos x)}{\sin x+\cos x} d x$
$= \frac{1}{2} x+\frac{1}{2} \int \frac{(\sin x-\cos x)}{\sin x+\cos x} d x$
Let sinx + cosx = t
$\Rightarrow \cos x-\sin x=\frac{d t}{d x}$
$\Rightarrow$ (cosx-sinx)dx = dt
$\Rightarrow -(\sin x-\cos x) dx = -dt$
Therefore, $I=\frac{x}{2}+\frac{1}{2} \int \frac{-d t}{t}$
$= \frac{x}{2}-\frac{1}{2} \log |t|+C$
$\Rightarrow I=\frac{x}{2}-\frac{1}{2} \log |\sin x+\cos x|+C$

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Question 44 Marks

By using the properties of definite integrals, evaluate the integral $\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$

Answer

Given $\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$
Let $I=\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$ .....(i)
as, $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-\mathrm{x}\right)\right] \mathrm{d} \mathrm{x}$
as $\left\{\tan (A-B)=\frac{\tan (A)-\tan (B)}{1+\tan (A) \tan (B)}\right\}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left[1+\frac{\tan \left(\frac{\pi}{4}\right)-\tan (\mathrm{x})}{1+\tan \left(\frac{\pi}{4}\right) \tan (\mathrm{x})}\right] \mathrm{dx}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left[1+\frac{1-\tan (\mathrm{x})}{1+\tan (\mathrm{x})}\right] \mathrm{d} \mathrm{x}$
$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \left[\frac{2}{1+\tan (x)}\right] d x$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log [2] \mathrm{d} \mathrm{x}-\int_{0}^{\frac{\pi}{4}} \log [1+\tan (\mathrm{x})] \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log [2] \mathrm{d} \mathrm{x}-\mathrm{I}$ (from (i))
$\Rightarrow 2 \mathrm{I}=[\mathrm{x} \log 2]_{0}^{\frac{\pi}{4}}$
$\Rightarrow 2 \mathrm{I}=\frac{\pi}{4} \log 2-0$
$\Rightarrow \mathrm{I}=\frac{\pi}{8} \log 2$

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Question 54 Marks

By using the properties of definite integrals, evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}$

Answer

Given integral is: $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$
Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$ .....(i)
as $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right.$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-\mathrm{x}\right)}{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-\mathrm{x}\right)+\cos ^{\frac{3}{2}}\left(\frac{\pi}{2}-\mathrm{x}\right)} \mathrm{dx}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{\frac{3}{2}} \mathrm{x}}{\cos ^{\frac{3}{2}} \mathrm{x}+\sin ^{\frac{3}{2}} \mathrm{x}} \mathrm{dx}$ ......(ii)
Adding (i) and (ii), we get
$2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$
$\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[1] \mathrm{d} \mathrm{x}$
$\Rightarrow 2 \mathrm{I}=[\mathrm{x}]_{0}^{\frac{\pi}{2}}$
$\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}-0$
$\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}$
$\Rightarrow \mathrm{I}=\frac{\pi}{4}$

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Question 64 Marks

By using the properties of definite integrals, evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

Answer

Given integral is: $\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$
Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ .....(i)
as $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\}$
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos \mathrm{x}}}{\sqrt{\cos \mathrm{x}}+\sqrt{\sin \mathrm{x}}} \mathrm{d} \mathrm{x}$ ......(ii)
Adding (i) and (ii), we get
$2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin \mathrm{x}}+\sqrt{\cos \mathrm{x}}}{\sqrt{\sin \mathrm{x}}+\sqrt{\cos \mathrm{x}}} \mathrm{dx}$
$\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[1] \mathrm{d} \mathrm{x}$
$\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}-0$
$\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}$
$\Rightarrow \mathrm{I}=\frac{\pi}{4}$

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Question 74 Marks

Show that $\int_0^a {f(x)g(x)dx = 2\int_0^a {f(x)dx} } $. If f and g are defined, $f(x) = f(a - x)$ and $g(x) + g(a - x) = 4$

Answer

$I = \int_0^a {f(x).g(x)dx} $

$ = \int_0^a {f(a - x).g(a - x)dx} $

$ = \int_0^a {f(x).[4 - g(x)]dx} $ [Using given condition]

$ = \int_0^a {4f(x)dx - \int_0^a {f(x).g(x)dx} } $

$I = 4\int_0^a {f(x)dx - I} $

$I = 2\int_0^a {f(x)dx} $

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Question 84 Marks

By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi} \log (1+\cos x) d x$

Answer

Given, $\int_{0}^{\pi} \log (1+\cos x) d x$
Let, $I=\int_{0}^{\pi} \log (1+\cos x) d x$ .....(i)
as, $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\}$
$\Rightarrow \mathrm{I}=\int_{0}^{\pi} \log (1+\cos (\pi-\mathrm{x}) \mathrm{d} \mathrm{x}$
$\Rightarrow I=\int_{0}^{\pi} \log (1-\cos x) d x$ ......(ii)
Adding (i) and (ii), we get
$2 \mathrm{I}=\int_{0}^{\pi}\{\log (1+\cos \mathrm{x})+\log (1-\cos \mathrm{x})\} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\int_{0}^{\pi} \log \left(1-\cos ^{2} \mathrm{x}\right) \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\int_{0}^{\pi} \log \left(\sin ^{2} \mathrm{x}\right) \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\int_{0}^{\pi} 2 \cdot \log (\sin \mathrm{x}) \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=2 \cdot \int_{0}^{\pi} \log (\sin \mathrm{x}) \mathrm{d} \mathrm{x}$
$I=\int_{0}^{\pi} \log (\sin x) d x$ ......(iii)
because, $\int_{0}^{2 a} f(x) d x=2 \cdot \int_{0}^{a} f(x) d x$ if f(2a - x) = f(x)
Here, if f(x) = log (sin x) and f($\pi$ - x) = log ( sin ($\pi$ - x))= log (sin x) = f(x)
$\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}} \log \sin \mathrm{xdx}$ ......(iv)
$\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-\mathrm{x}\right) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}} \log \cos \mathrm{x} \mathrm{d} \mathrm{x}$ ......(v)
Adding (1) and (2), we get
$\Rightarrow 2 \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}}(\log \sin \mathrm{x}+\log \cos \mathrm{x}) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(\log \sin \mathrm{x}+\log \cos \mathrm{x}+\log 2-\log 2) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(\log (2 \sin \mathrm{x} \cos \mathrm{x})-\log 2) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\left(\log (\sin 2 \mathrm{x}) \mathrm{d} \mathrm{x}-\int_{0}^{\frac{\pi}{2}} \log 2 \mathrm{d} \mathrm{x}\right.$
Let 2x = t $\Rightarrow$ 2dx = dt
When x = 0, t = 0 and when x = $\frac{\pi}{4}$, t = $\pi$
$\Rightarrow \mathrm{I}=\left[\frac{1}{2} \int_{0}^{\pi}(\log (\sin t) d t]-\left(\frac{\pi}{2} \log 2\right)\right.$
$\Rightarrow I=\left[\frac{I}{2}\right]-\left(\frac{\pi}{2} \log 2\right)$
$\Rightarrow \mathrm{I}=-\left(\frac{\pi}{2} \log 2\right)$
$\Rightarrow \mathrm{I}=-(\pi \log 2)$

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Question 94 Marks

By using the properties of definite integral, evaluate the integral $\int_0^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x$

Answer

According to the question , $ I=\int _ { 0 } ^ { \pi / 2 } ( 2 \log | \sin x | - \log | \sin 2 x | ) d x$
From $0 \ to \ {\pi\over2} , sinx \ and \ cosx$ are positive. so, $|sinx| = sinx , |cosx| = cosx$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } [ 2 \log ( \sin x ) - \log ( 2 \sin x \cos x ) ] d x$$[\because sin2x = 2sinx cosx]$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } [ 2 \log ( \sin x ) - ( \log 2 + \log ( \sin x )$$ + \log ( \cos x ) ) ] d x$$[\because log(ABC)= logA+logB+logC]$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } [ 2 \log ( \sin x ) - \log 2 - \log ( \sin x )$$ - \log ( \cos x ) ) ] d x$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } ( \log ( \sin x ) - \log 2 - \log ( \cos x ) ) d x$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } \log ( \sin x ) d x - \int _ { 0 } ^ { \pi / 2 } \log 2 d x$$ - \int _ { 0 } ^ { \pi / 2 } \log ( \cos x ) d x$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } \log \sin \left( \frac { \pi } { 2 } - x \right) d x - \log 2 [ x ] _ { 0 } ^ { \pi / 2 }$$ - \int _ { 0 } ^ { \pi / 2 } \log \cos x d x$ $[\because \int_b^axdx= \int_b^a(a+b-x)dx]$
$ \Rightarrow \quad I = \int _ { 0 } ^ { \pi / 2 } \log \cos x d x - \log 2 \left[ \frac { \pi } { 2 } - 0 \right]$$ - \int _ { 0 } ^ { \pi / 2 } \log \cos x d x$
$ \therefore \quad I = - \frac { \pi } { 2 } \log 2$

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Question 104 Marks

Evaluate $\int _ { - 1 } ^ { 2 } \left| x ^ { 3 } - x \right| d x$.

Answer

According to the question , $I =\int _ { - 1 } ^ { 2 } \left| x ^ { 3 } - x \right| d x$
We can observe that,
$\left| x ^ { 3 } - x \right| = \left\{ \begin{array} { c } { \left( x ^ { 3 } - x \right) , \text { when } - 1 < x < 0 } \\ { - \left( x ^ { 3 } - x \right) , \text { when } 0 \leq x < 1 } \\ { \left( x ^ { 3 } - x \right) , \text { when } 1 \leq x < 2 } \end{array} \right.$
By Splitting the intervals , we get
$I =\int _ { - 1 } ^ { 0 } \left| x ^ { 3 } - x \right| d x + \int _ { 0 } ^ { 1 } \left| x ^ { 3 } - x \right| d x +\int _ { 1} ^ { 2 } \left| x ^ { 3 } - x \right| d x$
$\therefore I = \int _ { - 1 } ^ { 0 } \left( x ^ { 3 } - x \right) d x + \int _ { 0 } ^ { 1 } - \left( x ^ { 3 } - x \right) d x + \int _ { 1 } ^ { 2 } \left( x ^ { 3 } - x \right) d x$
$= \int _ { - 1 } ^ { 0 } \left( x ^ { 3 } - x \right) d x - \int _ { 0 } ^ { 1 } \left( x ^ { 3 } - x \right) d x + \int _ { 1 } ^ { 2 } \left( x ^ { 3 } - x \right) d x$
$= \left[ \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right] _ { - 1 } ^ { 0 } - \left[ \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right] _ { 0 } ^ { 1 } + \left[ \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right] _ { 1 } ^ { 2 }$
$= \left[ 0 - \left( \frac { 1 } { 4 } - \frac { 1 } { 2 } \right) \right] - \left[ \left( \frac { 1 } { 4 } - \frac { 1 } { 2 } \right) - 0 \right]$$+ \left[ \left( \frac { 16 } { 4 } - \frac { 4 } { 2 } \right) - \left( \frac { 1 } { 4 } - \frac { 1 } { 2 } \right) \right]$
$= - \frac { 1 } { 4 } + \frac { 1 } { 2 } - \frac { 1 } { 4 } + \frac { 1 } { 2 } + 4 - 2 - \frac { 1 } { 4 } + \frac { 1 } { 2 }$
$= - \frac { 3 } { 4 } + \frac { 3 } { 2 } + 2$
$ = \frac { - 3 + 6 + 8 } { 4 }$
$ = \frac { 11 } { 4 }$
$\therefore I = \frac { 11 } { 4 }$ sq units.

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Question 114 Marks

Evalute $\int_{\frac{1}{3}}^1 \frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^4} d x$

Answer

$\therefore$ $I=\int_{1 / 3}^1 \frac{\left[x^3\left(\frac{1}{x^2}-1\right)\right]^{1 / 3}}{x^4} d x$
$\therefore$ $t=\int_{1 / 3}^1 \frac{x\left(\frac{1}{x^2}-1\right)^{1 / 3}}{x^4} d x$
$\therefore$ $=\int_{1 / 3}^1 \frac{\left(\frac{1}{x^2}-1\right)^{1 / 3}}{x^3} d x$
Let $\left(\frac{1}{x^2}-1\right)=t$
$\Longrightarrow-\frac{2}{x^3} d x=d t$
$\Longrightarrow \frac{d x}{x^3}=-\frac{d t}{2}$
Limits: $x=1 / 3 \rightarrow t=8, x=1 \rightarrow t=0$.
$I=\int_8^0 t^{1 / 3}\left(-\frac{d t}{2}\right)=\frac{1}{2} \int_0^8 t^{1 / 3} d t=\frac{1}{2}\left[\frac{3}{4} t^{4 / 3}\right]_0^8$
$I=\frac{3}{8}\left[8^{4 / 3}\right]=\frac{3}{8}[16]=6$

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Question 124 Marks

Evalute $\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$

Answer

Let $I=\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$...(i)
Using property $\int_0^a f(x) d x=\int_0^a f(a-x) d x$:
$\therefore$ $I=\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x$
= $\int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} d x$...(ii)
Adding (i) and (ii):
$\therefore$ $2 I=\int_0^\pi \frac{\pi \sin x}{1+\cos ^2 x} d x$
$\Longrightarrow 2 I=\pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x$
Let $\cos x=t$ $\Longrightarrow-\sin x d x=d t$.
Limits:$\therefore$ $x=0 \rightarrow t=1, x=\pi \rightarrow t=-1.2 I$
=$\pi \int_1^{-1} \frac{-d t}{1+t^2}=\pi \int_{-1}^1 \frac{d t}{1+t^2}=\pi\left[\tan ^{-1} t\right]_{-1}^1$
$\therefore$ $2 I=\pi\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right]=\pi\left[\frac{\pi}{2}\right]=\frac{\pi^2}{2} \Longrightarrow I =\frac{\pi^2}{4}$

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