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Maths 4 Marks

Integrals · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 33 questions.

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Question 14 Marks
Evaluate the integral $\int_{0}^{2} \frac{d x}{x+4-x^{2}}$ using substitution.
Answer
Given: $\int_{0}^{2} \frac{d x}{x+4-x^{2}}$ 
$\int_{0}^{2} \frac{d x}{x+4-x^{2}}=\int_{0}^{2} \frac{d x}{-\left(x^{2}-x-4\right)}$ 
We can write it as, $\int_{0}^{2} \frac{d x}{-\left(x^{2}-x+\frac{1}{4}-\frac{1}{4}-4\right)}$ 
= $\int_{0}^{2} \frac{d x}{-\left[\left(x-\frac{1}{2}\right)^{2}-\frac{17}{4}\right]}$ 
= $\int_{0}^{2} \frac{d x}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}\right]}$ 
Let $x-\frac{1}{2}=t \Rightarrow d x=d t$ 
When $x=0, t=-\frac{1}{2}$ and when $\mathrm{x}=2, \mathrm{t}=\frac{3}{2}$ 
$\Rightarrow \int_{0}^{2} \frac{d x}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}\right]}=\int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{d t}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-(t)^{2}\right]}$ 
because, $\int \frac{d x}{\left[(a)^{2}-(x)^{2}\right]}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C$ 
$\Rightarrow \int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{d t}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-(t)^{2}\right]}$ = $\left[\frac{1}{2\left(\frac{\sqrt{17}}{2}\right)} \log \frac{\left(\frac{\sqrt{17}}{2}+t\right)}{\frac{\sqrt{17}}{2}-t}\right]_{-\frac{1}{2}}^{\frac{3}{2}}$ 
= $\frac{1}{\sqrt{17}}\left[\log \frac{\left(\frac{\sqrt{17}}{2}+\frac{3}{2}\right)}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-\log \frac{\left(\frac{\sqrt{17}}{2}-\frac{1}{2}\right)}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right]$ 
= $\frac{1}{\sqrt{17}}\left[\log \frac{(\sqrt{17}+3)}{\sqrt{17}-3}-\log \frac{(\sqrt{17}-1)}{\sqrt{17}+1}\right]$ 
= $\frac{1}{\sqrt{17}}\left[\log \left\{\frac{(\sqrt{17}+3)}{\sqrt{17}-3} \times \frac{(\sqrt{17}+1)}{\sqrt{17}-1}\right\}\right]$ 
= $\frac{1}{\sqrt{17}}\left[\log \left\{\frac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)}\right\}\right]$ 
= $\frac{1}{\sqrt{17}} \log \left[\frac{17+3+4 \sqrt{17}}{17+3-4 \sqrt{17}}\right]$ 
= $\frac{1}{\sqrt{17}} \log \left[\frac{20+4 \sqrt{17}}{20-4 \sqrt{17}}\right]$ 
= $\frac{1}{\sqrt{17}} \log \left[\frac{5+\sqrt{17}}{5-\sqrt{17}}\right]$ 
= $\frac{1}{\sqrt{17}} \log \left[\frac{(5+\sqrt{17})(5+\sqrt{17})}{(5-\sqrt{17})(5+\sqrt{17})}\right]$ 
= $\frac{1}{\sqrt{17}} \log \left[\frac{(25+17+10 \sqrt{17})}{25-17}\right]$ 
= $\frac{1}{\sqrt{17}} \log \left[\frac{(42+10 \sqrt{17})}{8}\right]$ 
= $\frac{1}{\sqrt{17}} \log \left[\frac{(21+5 \sqrt{17})}{4}\right]$
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Question 24 Marks
Evaluate the integral $\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$ using substitution.
Answer
Given: $\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{x^{2}+1}\right) d x$ 
Let x = tan $\theta$ $\Rightarrow$ dx = sec2 $\theta$ d$\theta$ 
When, x = 0, $\theta$ = 0 and when x = 1, $\theta = \frac{\pi}{4}$ 
Let $I=\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{x^{2}+1}\right) d x$ 
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \sin ^{-1}\left(\frac{2 \tan \theta}{\tan ^{2} \theta+1}\right) \sec ^{2} \theta \mathrm{d} \theta$ 
= $\int_{0}^{\frac{\pi}{4}} \sin ^{-1}(\sin 2 \theta) \sec ^{2} \theta d \theta$ 
= $\int_{0}^{\frac{\pi}{4}} 2 \theta \sec ^{2} \theta d \theta$ 
= $2 \int_{0}^{\frac{\pi}{4}} \theta \sec ^{2} \theta d \theta$ 
By applying product rule as,
$\int \mathrm{u} \cdot \mathrm{v} \mathrm{d} \mathrm{x}=\mathrm{u} \cdot \int \mathrm{vdx}-\int \frac{\mathrm{du}}{\mathrm{dx}} \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=2\left[\theta \int \sec ^{2} \theta \mathrm{d} \theta-\int \frac{\mathrm{d}}{\mathrm{d} \theta} \theta \cdot\left\{\int \sec ^{2} \theta \mathrm{d} \theta\right\} \mathrm{d} \theta\right]_{0}^{\frac{\pi}{4}}$ 
= $2\left[\theta \tan \theta-\int 1 . \tan \theta d \theta\right]_{0}^{\frac{\pi}{4}}$ 
= $2[\theta \tan \theta-\log |\sec \theta|]_{0}^{\frac{\pi}{4}}$ 
= $2\left[\frac{\pi}{4} \tan \frac{\pi}{4}-\log \left|\sec \frac{\pi}{4}\right|-0+\log |\sec \theta|\right]$ 
= $2\left[\frac{\pi}{4}-\frac{1}{2} \log (2)\right.]$ 
= $\frac{\pi}{2}-\log (2)$ 
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Question 34 Marks
Evaluate the definite integral $\int _ { 0 } ^ { 1 } x e ^ { x ^ { 2 } } d x.$
Answer
Let $I = \int _ { 0 } ^ { 1 } x e ^ { x ^ { 2 } } d x$
Put $x ^ { 2 } = t \Rightarrow 2 x d x = d t \Rightarrow d x = \frac { d t } { 2 x }$
Lower limit when x = 0, then t = 0
Upper limit when x = 1, then t = 1.
$\therefore \quad I = \int _ { 0 } ^ { 1 } x e ^ { t } \frac { d t } { 2 x } = \frac { 1 } { 2 } \int _ { 0 } ^ { 1 } e ^ { t } d t$
$= \frac { 1 } { 2 } \left[ e ^ { t } \right] _ { 0 } ^ { 1 } = \frac { 1 } { 2 } \left[ e ^ { 1 } - e ^ { 0 } \right] = \frac { 1 } { 2 } [ e - 1 ]$
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Question 44 Marks
Integrate the function $\sqrt{1+3 x-x^{2}}$
Answer
$I=\int \sqrt{1+3 x-x^{2}} d x$ 
= $\int \sqrt{1-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)} d x$ 
= $\int \sqrt{\left(1+\frac{9}{4}\right)-\left(x-\frac{3}{2}\right)^{2}} d x$ 
= $\sqrt{\left(\frac{\sqrt{13}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}} d x$ 
We know that,
$\Rightarrow \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$ 
Therefore, 
$I=\frac{x-\frac{3}{2}}{2} \sqrt{1+3 x-x^{2}}+\frac{13}{4 \times 2} \sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}}\right)+C$ 
= $\frac{2 x-3}{4} \sqrt{1+3 x-x^{2}}+\frac{13}{8} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{13}}\right)+C$
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Question 54 Marks
Integrate the rational function $\frac{3 x+5}{x^{3}-x^{2}-x+1}$
Answer
We have, $\frac{3 x+5}{x^{3}-x^{2}-x+1}=\frac{3 x+5}{(x-1)^{2}(x+1)}$ 
Let $\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+1)}$ 
⇒3x + 5 = A(x - 1)(x + 1) + B(x + 1) + C( x - 1)2
⇒3x + 5 = A(x2 - 1) + B(x + 1) + C( x2 + 1 - 2x) …(i)
Substituting x = 1, in equation (i), we get,
B = 4
Equating the coefficients of x2 and x, we get,
A + C = 0
B – 2C = 3
$A=-\frac{1}{2}$ and $C=\frac{1}{2}$ 
Thus,
$\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}$ 
Now, $\int \frac{3 x+5}{(x-1)^{2}(x+1)} d x$ = $\int\left\{\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}\right\} d x$ 
= $\frac{-1}{2} \int \frac{1}{(x-1)} d x+4 \int \frac{1}{(x-1)^{2}} d x+\frac{1}{2} \int \frac{1}{(x+1)} d x$ 
= $-\frac{1}{2} \log |x-1|+4\left(\frac{-1}{x-1}\right)+\frac{1}{2} \log |x+1|+C$ 
= $\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C$ 
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Question 64 Marks
Integrate the rational function $\frac{x}{\left(x^{2}+1\right)(x-1)}$
Answer
Let  $\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C}{(x-1)}$ 
x = (Ax + B)(x - 1) + C (x2 + 1)
$\Rightarrow$ x = Ax2 - Ax + Bx - B + Cx2 + C
$\Rightarrow$ x = (A+C)x2 - (A-B)x - (B+C)
Equating the coefficients of x2, x and constant term, we get,
A + C = 0
-A + B = 1
-B + C = 0
On solving these equation, we get,
$A=-\frac{1}{2}, B=\frac{1}{2}$ and $C=\frac{1}{2}$ 
Thus, 
$\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{\left(-\frac{1}{2} x+\frac{1}{2}\right)}{\left(x^{2}+1\right)}+\frac{\frac{1}{2}}{(x-1)}$ 
$\Rightarrow~~\int \frac{x}{\left(x^{2}+1\right)(x-1)} d x=-\frac{1}{2} \int \frac{x}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x-1} d x$ 
= $-\frac{1}{4} \int \frac{2 x}{x^{2}+1} d x+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C$ 
Now, let us consider, $\int \frac{2 x}{x^{2}+1} d x$ Let (x2 + 1) = t
2xdx = dt
Thus,
$\int \frac{x}{\left(x^{2}+1\right)(x-1)} d x=-\frac{1}{4} \log \left|\mathrm{x}^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C$
= $\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+C$
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Question 74 Marks
Integrate the rational function $\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}$
Answer
We have, $\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=1-\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}$ 
Let $\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+3\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$ 
4x2 + 10 = (Ax + B)(x2 + 4) + (Cx + D)(x2 + 3)
⇒ 4x2 + 10 = Ax3 + 4Ax + Bx2 + 4B + Cx3 + 3Cx + Dx2 + 3D
⇒ 4x2 + 10 = (A+C)x3 + (B + D)x2 + (4A + 3C)x + (4B + 3D)
Equating the coefficients of x3, x2, x and constant term, we get,
A + C = 0
B + D = 0
4A + 3C = 0
4B + 3 B = 0
On solving these equations, we get,
A = 0, B = -2, C = 0 and D = 6
Therefore,
$\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{-2}{\left(x^{2}+3\right)}+\frac{6}{\left(x^{2}+4\right)}$ 
$\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=1-\left(\frac{-2}{\left(x^{2}+3\right)}+\frac{6}{\left(x^{2}+4\right)}\right)$ 
$\int \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)} d x=\int\left\{1+\frac{2}{\left(x^{2}+3\right)}-\frac{6}{\left(x^{2}+4\right)}\right\} d x$ 
= $\left\{1+\frac{2}{x^{2}+(\sqrt{3})^{2}}-\frac{6}{\left(x^{2}+2^{2}\right)}\right\}$ 
= $x+2\left(\frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}\right)-6\left(\frac{1}{2} \tan ^{-1} \frac{x}{2}\right)+C$ 
= $x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+C$ 
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Question 84 Marks
Integrate the rational function $\frac{1}{{{x^4} - 1}}$
Answer
$\frac{1}{{{x^4} - 1}}$

$= \frac{1}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}$

Putting ${x^2} = y$,we get,

$\frac{1}{{{x^4} - 1}}$$=\frac{1}{{\left( {y - 1} \right)\left( {y + 1} \right)}} $

$Let\ \frac{1}{{\left( {y - 1} \right)\left( {y + 1} \right)}} = \frac{A}{{y - 1}} + \frac{B}{{y + 1}}$ .....(i)

$\Rightarrow 1 = A\left( {y + 1} \right) + B\left( {y - 1} \right)$

$ \Rightarrow 1 = Ay + A + By - B$

Comparing the coefficients of y, A + B = 0 ......(ii)

Comparing constants A – B = 1 .......(iii)

On solving the eq. (ii) and (iii), we get $A = \frac{1}{2},B = \frac{{ - 1}}{2}$

Putting the values of A, B and y in eq. (i),

$\frac{1}{{{x^4} - 1}} = \frac{{\frac{1}{2}}}{{{x^2} - 1}} + \frac{{\frac{{ - 1}}{2}}}{{{x^2} + 1}}$

$\Rightarrow \int {\frac{1}{{{x^4} - 1}}dx = \frac{1}{2}\int {\frac{1}{{{x^2} - 1}}dx - \frac{1}{2}\int {\frac{1}{{{x^2} + 1}}dx} } } $

$= \frac{1}{2}.\frac{1}{{2.1}}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{1}{2}{\tan ^{ - 1}}x + c$

$= \frac{1}{4}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{1}{2}{\tan ^{ - 1}}x + c$

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Question 94 Marks
Integrate the rational function $\frac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}$
Answer
$\frac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}$

$= \frac{{2x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {2x + 3} \right)}}$

$ = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} + \frac{C}{{2x + 3}}$ ....(i)

$\Rightarrow $ 2x - 3 = A(x + 1)(2x + 3) + B(x - 1)(2x + 3) + C(x - 1)(x + 1)

$\Rightarrow $ 2x - 3 = A(2x2 + 5x + 3) + B(2x2 + x - 3) + C(x2 - 1)

$\Rightarrow $ 2x - 3 = 2Ax2 + 5Ax + 3A + 2Bx2 + Bx - 3B + Cx2 - C

Comparing coefficients of x2: 2A + 2B + C = 0 ...(ii)

Comparing coefficients of x : 5A + B = 2 .....(iii)

Comparing constants: 3A – 3B – C = –3 .....(iv)

On solving eq. (i), (ii) and (iii), we get $A = \frac{{ - 1}}{{10}},B = \frac{5}{2},C = \frac{{ - 24}}{5}$

Putting the values of A, B and C in eq. (i),

$\frac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}$

$= \frac{{\frac{{ - 1}}{{10}}}}{{x - 1}} + \frac{{\frac{5}{2}}}{{x + 1}} + \frac{{\frac{{ - 24}}{5}}}{{2x + 3}}$

$ \Rightarrow \int {\frac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}dx} $ $ = \frac{{ - 1}}{{10}}\int {\frac{1}{{x - 1}}dx + \frac{5}{2}} \int {\frac{1}{{x + 1}}dx - \frac{{24}}{5}\int {\frac{1}{{2x + 3}}dx} } $

$ = \frac{{ - 1}}{{10}}\log \left| {x - 1}+ \right|\frac{5}{2}\log \left| {x + 1} \right| - \frac{{24}}{5}\frac{{\log \left| {2x + 3} \right|}}{{2 }} + c$

$= \frac{{ - 1}}{{10}}\log \left| {x - 1}|+ \right|\frac{5}{2}\log \left| {x + 1} \right| - \frac{{12}}{5}\log \left| {2x + 3} \right| + c$

$= \frac{5}{2}\log \left| {x + 1} \right| - \frac{1}{{10}}\log \left| {x - 1} \right| - \frac{{12}}{5}\log \left| {2x + 3} \right| + c$

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Question 104 Marks
Integrate the function: $\frac{x-1}{\sqrt{x^{2}-1}}$
Answer
Here, $\int \frac{x-1}{\sqrt{x^{2}-1}} d x=\int \frac{x}{\sqrt{x^{2}-1}} d x-\int \frac{1}{\sqrt{x^{2}-1}} d x$
For $\int \frac{x}{\sqrt{x^{2}-1}} d x$ , Let x2 - 1 = t
$\Rightarrow$ 2x dx = dt
$\Rightarrow \int \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}=\frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}$ 
$= \frac{1}{2} \int t^{-\frac{1}{2}} d t$ 
$= \frac{1}{2}\left[2 t^{\frac{1}{2}}\right]=\sqrt{t}=\sqrt{x^{2}-1}$ 
$\Rightarrow \int \frac{\mathrm{x}-1}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}=\int \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}-\int \frac{1}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{d} \mathrm{x}$ 
$= \sqrt{x^{2}-1}- \log |x+\sqrt{x^{2}-1}|+C~~~$
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Question 114 Marks
Integrate the function $\frac{5 x+3}{\sqrt{x^{2}+4 x+10}}$
Answer
Let 5x + 3 = $A \frac{d}{d x}\left(x^{2}+4 x+10\right)+B$ 
$\Rightarrow$ 5x + 3 = A(2x + 4) + B
Now, equating the coefficients of x and constant term on both sides, we get,
2A = 5
$\Rightarrow A=\frac{5}{2}$ 
4A + B = 3
$\Rightarrow$ B = -7
$\Rightarrow 5 x+3=\frac{5}{2}(2 x+4)-7$ 
Again, $\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{\frac{5}{2}(2 x+4)-7}{\sqrt{x^{2}+4 x+10}} d x$ 
$\Rightarrow \frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$ 
Now, let us consider, $\int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$ 
Let x2 + 4x + 10 = t
⇒ (2x + 4) dx = dt
$\therefore \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{dt}{\sqrt t}=2 \sqrt{t}=2 \sqrt{x^{2}+4 x+10}$ ......(i)
And, Now let us consider, $\int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$ 
$\Rightarrow \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{1}{(\sqrt{x^{2}+4 x+4})+\sqrt6} d x$ 
$\Rightarrow \int \frac{1}{(x+2)^{2}+(\sqrt{6})^{2}} d x$ 
$=\log |(x+2) \sqrt{x^{2}+4 x+10}|$ .....(ii)
using eq. (i) and (ii), we get, 
$\Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\frac{5}{2}[2 \sqrt{x^{2}+4 x+10}]-7 \log (x+2) \sqrt{x^{2}+4 x+10} |+C$ 
$\Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=5 \sqrt{x^{2}+4 x+10}-7 \log |(x+2) \sqrt{x^{2}+4 x+10}|+C$
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Question 124 Marks
Integrate the function $\frac{x+3}{x^{2}-2 x-5}$
Answer
Let x + 3 = $A \frac{d}{d x}\left(x^{2}-2 x-5\right)+B$ 
$\Rightarrow$ x + 3 = A(2x - 2) + B
Now, equating the coefficients of x and constant term on both sides, we get,
2A = 1
$\Rightarrow A=\frac{1}{2}$ 
-2A + B = 3
$\Rightarrow$ B = 4
$\Rightarrow x+3=\frac{1}{2}(2 x-2)+4$ 
Now, $\int \frac{x+3}{x^{2}-2 x-5} d x=\int \frac{\frac{1}{2}(2 x-2)+4}{x^{2}-2 x-5} d x$ 
$=\frac{1}{2} \int \frac{2 x-2}{x^{2}-2 x-5} d x+4 \int \frac{1}{x^{2}-2 x-5} d x$ 
Now, Let us consider $\int \frac{2 x-2}{x^{2}-2 x-5} d x$ 
Let x2 - 2x - 5 = t
$\Rightarrow$ (2x - 2)dx = dt
$\therefore \int \frac{2 x-2}{x^{2}-2 x-5} d x=\int \frac{d t}{t}=\log |t|=\log \left|x^{2}-2 x-5\right|$ .......(i)
And, now let us consider, $\int \frac{1}{x^{2}-2 x-5} d x$ 
$\Rightarrow \int \frac{1}{\left(x^{2}-2 x+1\right)-6} d x$ 
$\Rightarrow \int \frac{1}{(x-1)^{2}+(\sqrt{6})^{2}} d x$ 
$=\frac{1}{2 \sqrt{6}} \log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right)$ ......(ii)
Using eq. (i) and (ii), we get,
$\Rightarrow \int \frac{x+3}{x^{2}-2 x-5} d x=\frac{1}{2} \log \left|x^{2}-2 x-5\right|+\frac{4}{2 \sqrt{6}} \log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right)+C$ 
$\Rightarrow \int \frac{x+3}{x^{2}-2 x-5} d x=\frac{1}{2} \log \left|x^{2}-2 x-5\right|+\frac{2}{\sqrt{6}} \log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right)+C$
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Question 134 Marks
Integrate the function $ \frac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}$
Answer
Let $I = \int {\frac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} $

Let Linear = A $\frac{d}{{dx}}$ (Quadratic) + B

$\Rightarrow x + 2 = A\frac{d}{{dx}}\left( {{x^2} + 2x + 3} \right) + B$

$ \Rightarrow $ x + 2 = A(2x + 2) + B …(ii)

$ \Rightarrow $ x + 2 = 2Ax + 2A + B

Comparing coefficients of x, 2A = 1 $ \Rightarrow A = \frac{1}{2}$

Comparing constants, 2A + B = 2

On solving, we get $A = \frac{1}{2},$ B = 1

Putting the values of A and B in eq. (ii),

$x + 2 = \frac{1}{2}\left( {2x + 2} \right) + 1$

Putting this value of x+2 in eq. (i),

$I = \int {\frac{{\frac{1}{2}\left( {2x + 2} \right) + 1}}{{\sqrt {{x^2} + 2x + 3} }}dx} $

$I = \frac{1}{2}\int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx + \int {\frac{1}{{\sqrt {{x^2} + 2x + 3} }}dx} } $

$\Rightarrow I = \frac{1}{2}{I_1} + {I_2}$ …(iii)

Now ${I_1} = \int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}} dx$

Putting x2 + 2x + 3 = t

$ \Rightarrow 2x + 2 = \frac{{dt}}{{dx}}$

$\Rightarrow $ (2x + 2)dx = dt

$\therefore {I_1} = \int {\frac{{dt}}{{\sqrt t }} = \int {{t^{\frac{{ - 1}}{2}}}dt} } $

$= \frac{{{t^{\frac{1}{2}}}}}{{\frac{1}{2}}} = 2\sqrt t $

$ = 2\sqrt {{x^2} + 2x + 3} $ …(iv)

Again ${I_2} = \int {\frac{1}{\sqrt{{x^2} + 2x + 3}}dx} $

$= \int {\frac{1}{{\sqrt {{x^2} + 2x + 1 + 2} }}} $

$= \int {\frac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} }}dx} $

$= \log \left| {x + 1 + \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} } \right|$

$= \log \left| {x + 1 + \sqrt {{x^2} + 2x + 3} } \right|$…(v)

Putting values of I1 and I2 in eq. (iii),

$I = \sqrt {{x^2} + 2x + 3} + \log \left| {x + 1 + \sqrt {{x^2} + 2x + 3} } \right| + c$

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Question 144 Marks
Integrate the function $\frac{x+2}{\sqrt{4 x-x^{2}}}$
Answer
Let x + 2 = $A \frac{d}{d x}\left(4 x-x^{2}\right)+B$ 
$\Rightarrow$ x + 2 = A(4 -2x) + B
Now, equating the coefficients of x and constant term on both sides, we get,
-2A = 1
$\Rightarrow A=-\frac{1}{2}$ 
and 4A + B = 2
$\Rightarrow$ B = 4
$\Rightarrow x+2=-\frac{1}{2}(4-2 x)+4$ 
Now, $\int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{4 x-x^{2}}} d x$ 
$\Rightarrow -\frac{1}{2} \int \frac{(4-2 x)}{\sqrt{4 x-x^{2}}} d x+4 \int \frac{1}{\sqrt{4 x-x^{2}}} d x$ 
Now, let us consider, $\int \frac{(4-2 x)}{\sqrt{4 x-x^{2}}} d x$ 
Let 4x - x2 = t
$\Rightarrow$ (4 - 2x) dx = dt
$\therefore \int \frac{(4-2 x)}{\sqrt{4 x-x^{2}}} d x=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}=2 \sqrt{4 x-x^{2}}$ .......(i)
And, Now let us consider, $\int \frac{1}{\sqrt{4 x-x^{2}}} d x$ 
Then, 4x - x2 = -(-4x + x2)
= (-4x + x2 + 4 - 4)
= 4 - (x - 2)2 
= (2)2 - (x - 2)2 
$\therefore \int \frac{1}{\sqrt{4 x-x^{2}}} d x=\sin ^{-1}\left(\frac{x-2}{2}\right)$ ......(ii)
using eq. (i) and (ii), we get,
$\Rightarrow \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=-\sqrt{4 x-x^{2}}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+C$
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Question 154 Marks
Integrate the function $\frac{6 x+7}{\sqrt{(x-5)(x-4)}}$
Answer
Let 6x + 7 = $A \frac{d}{d x}\left(x^{2}-9 x+20\right)+B$ 
$\Rightarrow$ 6x + 7 = A(2x - 9) + B
Now, equating the coefficients of x and constant term on both sides, we get,
2A = 6
$\Rightarrow$ A = 3
Also, -9A + B = 7
$\Rightarrow$ B = 34
$\Rightarrow$ 6x + 7 = 3 (2x - 9) + 34
$\therefore~ \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x$ 
$=\int \frac{3(2 x-9)+34}{\sqrt{x^{2}-9 x+20}} d x$
$= 3 \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x+34 \int \frac{1}{\sqrt{x^{2}-9 x+20}} d x$ 
Now, in $\int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x$ 
Let x2 - 9x + 20 = t
$\Rightarrow$ (2x - 9)dx = dt
$\therefore \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x=\int \frac{d t}{\sqrt{t}}$
$=2 \sqrt{t}$ 
$=2 \sqrt{x^{2}-9 x+20}$   ...(i)
And in $\int \frac{1}{\sqrt{x^{2}-9 x+20}} d x$, we have
x2 - 9x + 20 = $x^{2}-9 x+20+\frac{81}{4}-\frac{81}{4}$ 
$= \left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}$ 
$\Rightarrow \int \frac{1}{\sqrt{\mathrm{x}^{2}-9 \mathrm{x}+20}} \mathrm{dx}=\int \frac{1}{\sqrt{\left(\mathrm{x}-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} \mathrm{d} \mathrm{x}$ 
= $\log \left| (x-\frac{9}{2}\right)+\sqrt{x^{2}-9 x+20} |$ .....(ii)
Thus, from (i) and (ii), we get,
$ \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x=3[2 \sqrt{x^{2}-9 x+20}]+34\left[\log \left|\left(x-\frac{9}{2}\right)+\sqrt{x^{2}-9 x+20}\right|\right]+C$ 
$= 6 \sqrt{\mathrm{x}^{2}-9 \mathrm{x}+20}+34 \log \left[\left(\mathrm{x}-\frac{9}{2}\right)+\sqrt{\mathrm{x}^{2}-9 \mathrm{x}+20}\right]+\mathrm{C}$
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Question 164 Marks
Integrate the function $\frac{5 x-2}{1+2 x+3 x^{2}}$
Answer
Given integral is: $\frac{5 x-2}{1+2 x+3 x^{2}}$
Let 5x - 2 = $A \frac{d}{d x}\left(1+2 x+3 x^{2}\right)+B$ 
$\Rightarrow$ 5x - 2 = A(2 + 6x) + B
Now, equating the coefficients of x and constant term on both sides, we get,
6A = 5 $\Rightarrow A=\frac{5}{6}$
 2A + B = -2 $\Rightarrow B=-\frac{11}{3}$ 
$\Rightarrow 5 x-2=\frac{5}{6}(2+6 x)+\left(-\frac{11}{3}\right)$ 
$\therefore~ \int \frac{5 x+1}{1+2 x+3 x^{2}} d x=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^{2}} d x$ 
$= \frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^{2}} d x-\frac{11}{3} \int \frac{1}{1+2 x+3 x^{2}} d x$ 
Now, in $\int \frac{2+6 x}{1+2 x+3 x^{2}} d x$ 
Let 1 + 2x + 3x2 = t
$\Rightarrow$ (2 + 6x)dx = dt
$\therefore~\int \frac{2+6 \mathrm{x}}{1+2 \mathrm{x}+3 \mathrm{x}^{2}} \mathrm{dx}=\int \frac{\mathrm{dt}}{\mathrm{t}}=\log |\mathrm{t}|$ 
= log|1 + 2x + 3x2| ....…(1)
Also in $\int \frac{1}{1+2 x+3 x^{2}} d x$ 
$1+2 x+3 x^{2}=1+3\left(x^{2}+\frac{2}{3} x\right)$ 
$=3\left[\left(\mathrm{x}+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right]$ 
$\Rightarrow \int \frac{1}{1+2 x+3 x^{2}} d x=\frac{1}{3} \int \frac{1}{\left[\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right]} d x$ 
$= \frac{1}{3}\left[\frac{1}{\frac{\sqrt{2}}{3}} \tan ^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)\right]$ 
$= \frac{1}{3}\left[\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 \mathrm{x}+1}{\sqrt{2}}\right)\right]$ 
$=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)$ ......(ii)
Thus, from (1) and (2), we get,
$\Rightarrow \int \frac{5 x+1}{1+2 x+3 x^{2}} d x=\frac{5}{6}\left[\log \left|1+2 x+3 x^{2}\right|\right]-\frac{11}{3}\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)\right]+C$ 
$= \frac{5}{6}\left[\log \left|1+2 \mathrm{x}+3 \mathrm{x}^{2}\right|\right]-\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 \mathrm{x}+1}{\sqrt{2}}\right)+\mathrm{C}$
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Question 174 Marks
Integrate the function: $\frac{x+2}{\sqrt{x^{2}-1}}$
Answer
Clearly, we can write, x + 2 = $\frac{1}{2}(2 x)+2$
$\Rightarrow \int \frac{x+2}{\sqrt{x^{2}-1}} d x=\int \frac{\frac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} d x$ 
$= \frac{1}{2} \int \frac{2 \mathrm{x}}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}+2 \int \frac{1}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{d} \mathrm{x}$ 
Now in, $I_1$= $\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x$ 
Let x2 - 1 = t
$\Rightarrow$ (2x)dx = dt
$\Rightarrow I_1 = \frac{1}{2} \int \frac{2 \mathrm{x}}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}=\frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\frac{1}{2}[2 \sqrt{\mathrm{t}}]$ = $\frac{1}{2}[2 \sqrt{\mathrm{x^2-1}}]$ 
And $2 \int \frac{1}{\sqrt{x^{2}-1}} d x=2 \log |x+\sqrt{x^{2}-1}|$ 
$\Rightarrow \int \frac{\mathrm{x}+2}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}=\sqrt{\mathrm{x}^{2}-1}+2 \log |\mathrm{x}+\sqrt{\mathrm{x}^{2}-1}|+\mathrm{C}$
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Question 184 Marks
Integrate the function: $\frac{1}{\sqrt{(x-a)(x-b)}}$
Answer
We know that, (x - a)(x - b) can be written as x2 - (a + b)x + ab.
Then, x2 - (a + b)x + ab = x2 - (a + b)x + $\frac{(a+b)^{2}}{4}-\frac{(a+b)^{2}}{4}+a b$ 
$\Rightarrow \left[\mathrm{x}-\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)\right]^{2}-\frac{(\mathrm{a}-\mathrm{b})^{2}}{4}$ 
$\Rightarrow \int \frac{1}{\sqrt{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})}} \mathrm{dx}=\int \frac{1}{\sqrt{\left\{\mathrm{x}-\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)\right\}^{2}-\frac{(\mathrm{a}-\mathrm{b})^{2}}{4}} \mathrm{d} \mathrm{x}} \mathrm{dx}$ 
Let $x-\left(\frac{a+b}{2}\right)=t$ 
$\Rightarrow$ dx = dt
$\Rightarrow \int \frac{1}{\sqrt{\left\{x-\left(\frac{a+b}{2}\right)\right\}^{2}-\frac{(a-b)^{2}}{4}}} d x=\int \frac{1}{\sqrt{t^{2}-\frac{(a-b)^{2}}{4}}}$ 
$= \log |t+\sqrt{t^{2}-\frac{(a-b)^{2}}{4}}|+C$ 
$=\log \left|\left\{\mathrm{x}-\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)\right\}+\sqrt{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})}\right|+\mathrm{C}$
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Question 194 Marks
Integrate the function $\frac{1}{{\sqrt {8 + 3x - {x^2}} }}$
Answer
$\int {\frac{1}{{\sqrt {8 + 3x - {x^2}} }}} dx$ $ = \int {\frac{1}{{\sqrt { - {x^2} + 3x + 8} }}} dx$
$= \int {\frac{1}{{\sqrt { - \left( {{x^2} - 3x - 8} \right)} }}} dx$
$= \int {\frac{1}{{\sqrt { - \left\{ {{x^2} - 3x + {{\left( {\frac{3}{2}} \right)}^2} - {{\left( {\frac{3}{2}} \right)}^2} - 8} \right\}} }}dx} $
$ = \int {\frac{1}{{\sqrt { - \left\{ {{{\left( {x - \frac{3}{2}} \right)}^2} - \frac{{41}}{4}} \right\}} }}dx} $
$= \int {\frac{1}{{\sqrt {{{\left( {\frac{{\sqrt {41} }}{2}} \right)}^2} - {{\left( {x - \frac{3}{2}} \right)}^2}} }}dx} $
$= {\sin ^{ - 1}}\frac{{x - \frac{3}{2}}}{{\frac{{\sqrt {41} }}{2}}} + c$
$= {\sin ^{ - 1}}\left( {\frac{{2x - 3}}{{\sqrt {41} }}} \right) + c$
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Question 204 Marks
Integrate the function $\frac{1}{\sqrt{(x-1)(x-2)}}$
Answer
Clearly, $\int \frac{1}{\sqrt{(x-1)(x-2)}} d x=\int \frac{1}{\sqrt{x^{2}-3 x+2}} d x=\int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} d x$ 
Let $x-\frac{3}{2}=t$ 
$\Rightarrow$ dx = dt
$\Rightarrow \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} d x=\int \frac{1}{\sqrt{(t)^{2}-\left(\frac{1}{2}\right)^{2}}} d t$ 
$= \log |t+\sqrt{(t)^{2}-\left(\frac{1}{2}\right)^{2}}|+C$ 
$= \log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^{2}-3 x+2}\right|+C$
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Question 214 Marks
Integrate the function: $\frac{1}{9 x^{2}+6 x+5}$
Answer
Clearly, $9 x^{2}+6 x+5$ = $(3 x+1)^{2}+(2)^{2}$
$\Rightarrow \int \frac{1}{9 x^{2}+6 x+5} d x=\int \frac{1}{(3 x+1)^{2}+(2)^{2}} d x$ 
Let 3x + 1 = t
$\Rightarrow$ 3dx = dt
$\therefore~~ \int \frac{1}{(3 x+1)^{2}+(2)^{2}} d x=\frac{1}{3} \int \frac{1}{t^{2}+2^{2}} d t$ 
$= \frac{1}{3}\left[\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)\right]+C$ 
$= \frac{1}{6} \tan ^{-1}\left(\frac{3 \mathrm{x}+1}{2}\right)+\mathrm{C}$
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Question 224 Marks
Find the integral of the function sin3 (2x + 1)
Answer
Let $I=\int \sin ^{3}(2 x+1) d x$
$= \int \sin ^{3}(2 x+1) d x=\int \sin ^{2}(2 x+1) \cdot \sin (2 x+1) d x$ 
$= \int\left(1-\cos ^{2}(2 x+1)\right) \sin (2 x+1) d x$ 
Let cos (2x + 1) = t
$\Rightarrow$ -2sin(2x + 1)dx = dt
$\Rightarrow$ sin(2x + 1)dx = $\frac{-\mathrm{dt}}{2}$ 
$\therefore~ I=\frac{-1}{2} \int\left(1-t^{2}\right) d t$ 
$= \frac{-1}{2}\left\{t-\frac{t^{3}}{3}\right\}$ 
$= \frac{-1}{2}\left\{\cos (2 \mathrm{x}+1)-\frac{\cos ^{3}(2 \mathrm{x}+1)}{3}\right\}$ 
$= \frac{-\cos (2 \mathrm{x}+1)}{2}+\frac{\cos ^{3}(2 \mathrm{x}+1)}{6}+\mathrm{C}$
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Question 234 Marks
Find the integrals of the function cos 2x cos 4x cos 6x
Answer
Clearly,
$\int \cos 2 x \cos 4 x \cos 6 x d x$ = $\int \cos 2 x\left[\frac{1}{2}\{\cos (4 x+6 x)+\cos (4 x-6 x)\}\right] d x$ 
$=\frac{1}{2} \int\{\cos 2 x \cos 10 x+\cos 2 x \cos (-2 x)\} d x$ 
$= \frac{1}{2} \int\left\{\cos 2 x \cos 10 x+\cos ^{2} 2 x\right\} d x$ 
$= \frac{1}{2} \int\left[\frac{1}{2} \left\{\cos (2 \mathrm{x}+10 \mathrm{x})+\cos (2 \mathrm{x}-10 \mathrm{x})\right\}+\left(\frac{1+\cos 4 \mathrm{x}}{2}\right)\right] \mathrm{dx}$ 
$= \frac{1}{4} \int(\cos 12 x+\cos 8 x+1+\cos 4 x) d x$ 
$= \frac{1}{4}\left[\frac{\sin 12 \mathrm{x}}{12}+\frac{\sin 8 \mathrm{x}}{8}+\mathrm{x}+\frac{\sin 4 \mathrm{x}}{4}\right]+\mathrm{c}$
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Question 244 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 254 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 264 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 274 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 284 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 294 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 304 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 314 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 324 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]$$ - \int _ { 0 } ^ { \pi / 2 } \log \cos x d x$
$ \therefore \quad I = - \frac { \pi } { 2 } \log 2$
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Question 334 Marks
Evaluate $\int _ { - 1 } ^ { 2 } \left| x ^ { 3 } - x \right| d x$.
Answer
According to the question , $I =\int _ { - 1 } ^ { 2 } \left| x ^ { 3 } - x \right| d x$
We can observe that,
$\left| x ^ { 3 } - x \right| = \left\{ \begin{array} { c } { \left( x ^ { 3 } - x \right) , \text { when } - 1 < x < 0 } \\ { - \left( x ^ { 3 } - x \right) , \text { when } 0 \leq x < 1 } \\ { \left( x ^ { 3 } - x \right) , \text { when } 1 \leq x < 2 } \end{array} \right.$
By Splitting the intervals , we get  
$I =\int _ { - 1 } ^ { 0 } \left| x ^ { 3 } - x \right| d x + \int _ { 0 } ^ { 1 } \left| x ^ { 3 } - x \right| d x +\int _ { 1} ^ { 2 } \left| x ^ { 3 } - x \right| d x$
 $\therefore I = \int _ { - 1 } ^ { 0 } \left( x ^ { 3 } - x \right) d x + \int _ { 0 } ^ { 1 } - \left( x ^ { 3 } - x \right) d x + \int _ { 1 } ^ { 2 } \left( x ^ { 3 } - x \right) d x$
$= \int _ { - 1 } ^ { 0 } \left( x ^ { 3 } - x \right) d x - \int _ { 0 } ^ { 1 } \left( x ^ { 3 } - x \right) d x + \int _ { 1 } ^ { 2 } \left( x ^ { 3 } - x \right) d x$
$= \left[ \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right] _ { - 1 } ^ { 0 } - \left[ \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right] _ { 0 } ^ { 1 } + \left[ \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right] _ { 1 } ^ { 2 }$
$= \left[ 0 - \left( \frac { 1 } { 4 } - \frac { 1 } { 2 } \right) \right] - \left[ \left( \frac { 1 } { 4 } - \frac { 1 } { 2 } \right) - 0 \right]$$+ \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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4 Marks - Maths STD 12 Science Questions - Vidyadip