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Maths Solve the Following Question.(5 Marks)

Indefinite Integration · Maharashtra Board STD 12 Science (MSBSHSE - English Medium). 47 questions.

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Solve the Following Question.(5 Marks)

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Question 15 Marks
Integrate the following with respect to the respective variable:

$\frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x}$

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Question 75 Marks
Evaluate : $\int \frac{\tan \theta+\tan ^3 \theta}{1+\tan ^3 \theta} \cdot d \theta$
Answer
$\mathrm{I}=\int \frac{(\tan \theta)\left(1+\tan ^2 \theta\right)}{1+\tan ^3 \theta} \cdot d \theta \quad=\int \frac{(\tan \theta) \cdot\left(1+\tan ^2 \theta\right)}{1+\tan ^3 \theta} \cdot d \theta=\int \frac{\tan \theta \cdot \sec ^2 \theta}{1+\tan ^3 \theta} \cdot d \theta$ put $\tan \theta=x \quad \therefore \quad \sec ^2 \theta \cdot d \theta=1 \cdot d x$ $=\int \frac{x}{1+x^3} \cdot d x=\int \frac{x}{(1+x)\left(1-x+x^2\right)} \cdot d x$
$
\text { Consider, } \begin{aligned}
\frac{x}{(1+x)\left(1-x+x^2\right)} & =\frac{A}{1+x}+\frac{B x+C}{\left(1-x+x^2\right)} \\
& =\frac{A\left(1-x+x^2\right)+B x+C(1+x)}{(1+x)\left(1-x+x^2\right)} \\
\therefore x & =A\left(1-x+x^2\right)+(B x+C)(1+x)=A-A x+A x^2+B x+B x^2+C+C x \\
0 x^2+1 \cdot x+0 & =(A+B) x^2+(-A+B+C) x+(A+C)
\end{aligned}
$
compairing the co-efficients of like powers of variables.
$0=A+B$
$1=-A+B+C$
and
$0=A+C$
Solving these equations, we get $A=-\frac{1}{3} ; B=\frac{1}{3}$ and $C=\frac{1}{3}$
Thus,
$
\frac{x}{(1+x)\left(1-x+x^2\right)}=\frac{\left(-\frac{1}{3}\right)}{1+x}+\frac{\left(\frac{1}{3} x+\frac{1}{3}\right)}{\left(1-x+x^2\right)}
$
$
\begin{aligned}
\therefore \mathrm{I} & =\int\left[\frac{\left(-\frac{1}{3}\right)}{1+x}+\frac{\left(\frac{1}{3} x+\frac{1}{3}\right)}{\left(1-x+x^2\right)}\right] \cdot d x=-\frac{1}{3} \cdot \int \frac{1}{1+x} \cdot d x+\frac{1}{3} \cdot \int \frac{x+1}{1-x+x^2} \cdot d x \\
& =-\frac{1}{3} \cdot \int \frac{1}{1+x} \cdot d x+\frac{1}{3} \cdot \frac{1}{(2)} \int \frac{2 x-1+3}{x^2-x+1} \cdot d x \quad \because \quad \frac{d}{d x} x^2-x+1=2 x-1 \\
& =-\frac{1}{3} \cdot \int \frac{1}{1+x} \cdot d x+\frac{1}{3} \cdot \frac{1}{2} \cdot \int \frac{2 x-1+3}{x^2-x+1} \cdot d x \\
& =-\frac{1}{3} \cdot \int \frac{1}{1+x} \cdot d x+\frac{1}{6} \cdot \int \frac{2 x-1}{x^2-x+1} \cdot d x+\frac{1}{6} \cdot \int \frac{3}{x^2-x+1} \cdot d x \\
& =\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3
\end{aligned}
$

$\begin{aligned} & \therefore \quad \mathrm{I}_1=-\frac{1}{3} \cdot \int \frac{1}{1+x} \cdot d x \quad=-\frac{1}{3}[\log (1+x)] \\ & =-\frac{1}{3} \log (1+\tan \theta) \\ & \therefore \quad \mathrm{I}_2=\frac{1}{6} \cdot \int \frac{2 x-1}{x^2-x+1} \cdot d x=\frac{1}{6}\left[\log \left(x^2-x+1\right)\right] \\ & =\frac{1}{6} \log \left(\tan ^2 \theta-\tan \theta+1\right) \\ & \therefore \quad \mathrm{I}_3=\frac{1}{6} \cdot \int \frac{3}{x^2-x+1} \cdot d x \\ & =\frac{1}{2} \cdot \int \frac{1}{x^2-x+\frac{1}{4}-\frac{1}{4}+1} \cdot d x \quad \because \quad\left\{\left(\frac{1}{2} \text { coefficient of } x\right)^2=\left(\frac{1}{2}(-1)\right)^2=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\right\} \\ & =\frac{1}{2} \cdot \int \frac{1}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \cdot d x \\ & =\frac{1}{2} \cdot\left[\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}\right] \cdot \tan ^{-1}\left[\frac{x-\frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)}\right]+c \\ & =\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c \\ & \therefore \quad \mathrm{I}_3=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan \theta-1}{\sqrt{3}}\right)+c \\ & \therefore \quad \int \frac{\tan \theta+\tan ^3 \theta}{1+\tan ^3 \theta} \cdot d \theta=-\frac{1}{3} \log (1+\tan \theta)+\frac{1}{6} \log \left(\tan ^2 \theta-\tan \theta+1\right)+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan \theta-1}{\sqrt{3}}\right)+c \\ & \end{aligned}$

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Question 85 Marks
Evaluate : $\int \frac{1}{2 \cos x+\sin 2 x} \cdot d x$
Answer
$
\begin{aligned}
& \mathrm{I}=\int \frac{1}{2 \cos x+\sin 2 x} \cdot d x \quad=\int \frac{1}{2 \cos x+2 \sin x \cdot \cos x} \cdot d x=\int \frac{1}{2(\cos x)(1+\sin x)} \cdot d x \\
& =\frac{1}{2} \cdot \int \frac{\cos x}{\cos ^2 x(1+\sin x)} \cdot d x=\frac{1}{2} \cdot \int \frac{\cos x}{\left(1-\sin ^2 x\right)(1+\sin x)} \cdot d x \\
&
\end{aligned}
$
put $\sin x=t$
$
\therefore \quad \cos x \cdot d x=1 \cdot d t
$
$
=\frac{1}{2} \cdot \int \frac{1}{\left(1-t^2\right)(1+t)} \cdot d t=\frac{1}{2} \cdot \int \frac{1}{(1-t)(1+t)(1+t)} \cdot d t=\frac{1}{2} \cdot \int \frac{1}{(1-t)(1+t)^2} \cdot d t
$
Consider, $\quad \frac{1}{(1-t)(1+t)^2}=\frac{A}{(1-t)}+\frac{B}{(1+t)}+\frac{C}{(1+t)^2}=\frac{A(1+t)^2+B(1-t)(1+t)+C(1-t)}{(1-t)(1+t)^2}$
$
\therefore \quad 1=A(1+t)^2+B(1-t)(1+t)+C(1-t)
$
at $t=1$,
$1=A(2)^2+B(0)+C(0)$
$1=4 A \quad \Rightarrow A=\frac{1}{4}$
at $t=-1, \quad 1=A(0)+B(0)+C(2)$
$1=2 C \quad \Rightarrow C=\frac{1}{2}$
at $t=0, \quad 1=A(1)^2+B(1)(1)+C(1)$
$1=A+B+C$
$1=\frac{1}{4}+B+\frac{1}{2} \Rightarrow B=\frac{1}{4}$
Thus, $\quad \frac{1}{(1-t)(1+t)^2}=\frac{\left(\frac{1}{4}\right)}{(1-t)}+\frac{\left(\frac{1}{4}\right)}{(1+t)}+\frac{\left(\frac{1}{2}\right)}{(1+t)^2}$
$\therefore \mathrm{I}=\int\left[\frac{\left(\frac{1}{4}\right)}{(1-t)}+\frac{\left(\frac{1}{4}\right)}{(1+t)}+\frac{\left(\frac{1}{2}\right)}{(1+t)^2}\right] \cdot d t=\frac{1}{2}\left[\frac{1}{4} \log (1-t) \cdot \frac{1}{(-1)}+\frac{1}{4} \log (1+t)+\frac{1}{2} \cdot \frac{(-1)}{(1+t)}\right]+c$
$=\frac{1}{2}\left[\frac{1}{4} \log (1-t) \cdot \frac{1}{(-1)}+\frac{1}{4} \log (1+t)+\frac{1}{2} \cdot \frac{-1}{1+t}\right]+c$
$=\frac{1}{8}\left[-\log (1-\sin x)+\log (1+\sin x)-\frac{2}{1+\sin x}\right]+c=\frac{1}{8}\left[\log \left(\frac{1+\sin x}{1-\sin x}\right)-\frac{2}{1+\sin x}\right]+c$
$\therefore \quad \int \frac{1}{2 \cos x+\sin 2 x} \cdot d x=\frac{1}{8}\left[\log \left(\frac{1+\sin x}{1-\sin x}\right)-\frac{2}{1+\sin x}\right]+c$
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Question 95 Marks
Evaluate : $\int \frac{1}{(\sin \theta)(3+2 \cos \theta)} \cdot d \theta$
Answer
$
\begin{aligned}
& \mathrm{I}=\int \frac{1}{(\sin \theta)(3+2 \cos \theta)} \cdot d \theta=\int \frac{\sin \theta}{\left(1-\cos ^2 \theta\right)(3+2 \cos \theta)} \cdot d \theta \\
& =\int \frac{\sin \theta}{(1-\cos \theta)(1+\cos \theta)(3+2 \cos \theta)} \cdot d \theta \\
& \text { put } \cos \theta=t \quad \therefore \quad-\sin \theta \cdot d \theta=1 \cdot d t \\
& \therefore \sin \theta \cdot d \theta=-1 \cdot d t \\
&
\end{aligned}
$
$
\text { Consider, } \begin{aligned}
\frac{-1}{(1-t)(1+t)(3+2 t)} & =\frac{A}{(1-t)}+\frac{B}{(1+t)}+\frac{C}{(3+2 t)} \\
& =\frac{A(1+t)(3+2 t)+B(1-t)(3+2 t)+C(1-t)(1+t)}{(1-t)(1+t)(3+2 t)}
\end{aligned}
$
$
\therefore \quad-1=A(1+t)(3+2 t)+B(1-t)(3+2 t)+C(1-t)(1+t)
$
at $t=1$,
$
\begin{aligned}
& -1=A(2)(5)+B(0)+C(0) \\
& -1=10 A \Rightarrow A=-\frac{1}{10}
\end{aligned}
$
at $t=-1$,
$
\begin{aligned}
& -1=A(0)+B(2)(1)+C(0) \\
& -1=2 B \quad \Rightarrow \quad B=-\frac{1}{2}
\end{aligned}
$
at $t=-\frac{3}{2}, \quad-1=A(0)+B(0)+C\left(+\frac{5}{2}\right)\left(-\frac{1}{2}\right)$
$
-1=-\frac{5}{4} C \quad \Rightarrow C=\frac{4}{5}
$
Thus,
$
\frac{-1}{(1-t)(1+t)(3+2 t)}=\frac{\left(-\frac{1}{10}\right)}{(1-t)}+\frac{\left(-\frac{1}{2}\right)}{(1+t)}+\frac{\left(\frac{4}{5}\right)}{(3+2 t)}
$
$
\begin{aligned}
\therefore \mathrm{I} & =\int\left[\frac{\left(-\frac{1}{10}\right)}{(1-t)}+\frac{\left(-\frac{1}{2}\right)}{(1+t)}+\frac{\left(\frac{4}{5}\right)}{(3+2 t)}\right] \cdot d t \\
& =-\frac{1}{10} \log (1-t) \cdot \frac{1}{(-1)}-\frac{1}{2} \log (1+t)+\frac{4}{5} \log (3+2 t) \cdot \frac{1}{2}+c \\
& =\frac{1}{10} \log (1-\cos \theta)-\frac{1}{2} \log (1+\cos \theta)+\frac{4}{10} \log (3+2 \cos \theta)+c \\
& =\frac{1}{10}\left(\log \frac{(1-\cos \theta)(3+2 \cos \theta)^4}{(1+\cos \theta)^5}\right)+c \quad \because \log a^m=m \cdot \log a
\end{aligned}
$
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Question 105 Marks
Evaluate : $\int \frac{2 x^2-3}{\left(x^2-5\right)\left(x^2+4\right)} \cdot d x$
Answer
Consider, $\frac{2 x^2-3}{\left(x^2-5\right)\left(x^2+4\right)}$
Let
$
\begin{array}{ll}
\text { Let } & x^2=m \\
\therefore & =\frac{2 m-3}{(m-5)(m+4)} \ldots \text { proper rational function. }
\end{array}
$
Now, $\frac{2 m-3}{(m-5)(m+4)}=\frac{A}{(m-5)}+\frac{B}{(m+4)}=\frac{A(m+4)+B(m-5)}{(m-5)(m+4)}$
$
\begin{aligned}
& \therefore \quad 2 m-3=A(m+4)+B(m-5) \\
& \text { at } m=5, \quad 2(5)-3 \quad=A(9)+B(0) \\
& 7=9 A \quad \Rightarrow A=\frac{7}{9} \\
& \text { at } m=-4, \quad 2(-4)-3=A(0)+B(-9) \\
& -11=-9 B \Rightarrow B=\frac{11}{9}
\end{aligned}
$
Thus, $\frac{2 m-3}{(m-5)(m+4)}=\frac{\left(\frac{7}{9}\right)}{(m-5)}+\frac{\left(\frac{11}{9}\right)}{(m+4)} \quad$ i.e. $\frac{2 x^2-3}{\left(x^2-5\right)\left(x^2+4\right)}=\frac{\left(\frac{7}{9}\right)}{x^2-5}+\frac{\left(\frac{11}{9}\right)}{x^2+4}$
$
\begin{aligned}
& \therefore \mathrm{I}=\int\left[\frac{\left(\frac{7}{9}\right)}{x^2-5}+\frac{\left(\frac{11}{9}\right)}{x^2+4}\right] \cdot d x \\
& =\frac{7}{9} \cdot \int \frac{1}{x^2-(\sqrt{5})^2} \cdot d x+\frac{11}{9} \cdot \int \frac{1}{x^2+(2)^2} \cdot d x \\
& =\frac{7}{9} \cdot \frac{1}{2(\sqrt{5})} \cdot \log \left[\frac{x-\sqrt{5}}{x+\sqrt{5}}\right]+\frac{11}{9} \cdot \frac{1}{2} \cdot \tan ^{-1}\left(\frac{x}{2}\right)+c \\
& \therefore \mathrm{I}=\frac{7}{18(\sqrt{5})} \cdot \log \left[\frac{x-\sqrt{5}}{x+\sqrt{5}}\right]+\frac{11}{18} \cdot \tan ^{-1}\left(\frac{x}{2}\right)+c \\
& \therefore \int \frac{2 x^2-3}{\left(x^2-5\right)\left(x^2+4\right)} \cdot d x=\frac{7}{18(\sqrt{5})} \cdot \log \left[\frac{x-\sqrt{5}}{x+\sqrt{5}}\right]+\frac{11}{18} \cdot \tan ^{-1}\left(\frac{x}{2}\right)+c \\
&
\end{aligned}
$
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Question 125 Marks
Evaluate : $\int \sqrt{\frac{x-5}{x-7}} \cdot d x$
Answer
$\mathrm{I}=\int \sqrt{\frac{(x-5) \cdot(x-5)}{(x-7) \cdot(x-5)}} \cdot d x=\int \sqrt{\frac{(x-5)^2}{x^2-12 x+35}} \cdot d x$
$
\because \quad \begin{aligned}
x-5 & =A \cdot \frac{d}{d x}\left(x^2-12 x+35\right)+B \\
x-5 & =A(2 x-12)+B \\
& =(2 A) x+(-12 A+B)
\end{aligned}
$
compairing, the co-efficients of like variables and constants
$
\begin{aligned}a
& 2 A=1 \text { and }-12 A+B=-5 \\
\Rightarrow & A=\frac{1}{2} \text { and } B=1
\end{aligned}
$
$\begin{aligned} & \mathrm{I}=\int \frac{\frac{1}{2} \cdot \frac{d}{d x}\left(x^2-12 x+35\right)+(1)}{\sqrt{x^2-12 x+35}} \cdot d x \\ & =\frac{1}{2} \cdot \int \frac{\frac{d}{d x}\left(x^2-12 x+35\right)}{\sqrt{x^2-12 x+35}} \cdot d x+\int \frac{1}{\sqrt{x^2-12 x+35}} \cdot d x \\ & =\mathrm{I}_1+\mathrm{I}_2 \\ & \therefore \quad I_1=\frac{1}{2} \cdot \int \frac{2 x-12}{\sqrt{x^2-12 x+35}} \cdot d x \\ & \text { put } x^2-12 x+35=t \\ & \therefore(2 x-12) \cdot d x=1 \cdot d t \\ & \mathrm{I}_1 \quad=\frac{1}{2} \cdot \int \frac{1}{\sqrt{t}} \cdot d t \\ & =\int \frac{1}{2 \sqrt{t}} \cdot d t \\ & =\sqrt{t}+c_1 \\ & =\sqrt{x^2-12 x+35}+c \\ & \end{aligned}$
$
\begin{aligned}
& \therefore \quad \mathrm{I}_2=\int \frac{1}{\sqrt{x^2-12 x+35}} \cdot d x \\
& =\int \frac{1}{\sqrt{x^2-12 x+36-1}} \cdot d x \\
& =\int \frac{1}{\sqrt{(x-6)^2-(1)^2}} \cdot d x \\
& \because \quad \int \frac{1}{\sqrt{X^2-A^2}} \cdot d x=\log \left(X+\sqrt{X^2-A^2}\right)+c \\
& \mathrm{I}_2 \quad=\log \left((x-6)+\sqrt{(x-6)^2-1}\right)+c_2 \\
& =\log \left((x-6)+\sqrt{x^2-12 x+35}\right)+c_2 \\
&
\end{aligned}
$
Thus, from (i), (ii) and (iii)
$
\begin{aligned}
& \int \sqrt{\frac{x-5}{x-7}} \cdot d x \\
= & \sqrt{x^2-12 x+35}+\log \left((x-6)+\sqrt{x^2-12 x+35}\right)+c \\
& \left(c_1+c_2=c\right)
\end{aligned}
$
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Question 135 Marks
Evaluate : $\int \frac{2 x-3}{3 x^2+4 x+5} \cdot d x$
Answer
$\quad 2 x-3=A \cdot \frac{d}{d x}\left(3 x^2+4 x+5\right)+B$
$
\begin{aligned}
2 x-3 & =A(6 x+4)+B \\
& =(6 A) x+(4 A+B)
\end{aligned}
$
compairing the sides/ the co-efficients of like variables and constants
$
\begin{aligned}
& 6 A=2 \text { and } 4 A+B=-3 \\
& \Rightarrow A=\frac{1}{3} \text { and } B=-\frac{13}{3} \\
& =\int \frac{\frac{1}{3} \cdot \frac{d}{d x}\left(3 x^2+4 x+5\right)+\left(-\frac{13}{3}\right)}{3 x^2+4 x+5} \cdot d x \\
& =\frac{1}{3} \cdot \int \frac{\frac{d}{d x}\left(3 x^2+4 x+5\right)}{3 x^2+4 x+5} \cdot d x-\frac{13}{3} \int \frac{1}{3 x^2+4 x+5} \cdot d x \\
& =\frac{1}{3} \cdot \int \frac{6 x+4}{3 x^2+4 x+5} \cdot d x-\frac{13}{3} \int \frac{1}{3 x^2+4 x+5} \cdot d x \\
& =\mathrm{I}_1-\mathrm{I}_2 \\
& \therefore \quad \mathrm{I}_1=\frac{1}{3} \cdot \int \frac{6 x+4}{3 x^2+4 x+5} \cdot d x \\
& \text { put } 3 x^2+4 x+5=t \\
& \therefore \quad(6 x+4) \cdot d x=1 \cdot d t \\
& \mathrm{I}_1=\frac{1}{3} \cdot \int \frac{1}{t} \cdot d t \\
& =\frac{1}{3} \cdot \log (t)+c_1 \\
& =\frac{1}{3} \cdot \log \left(3 x^2+4 x+5\right)+c_1 \\
\end{aligned}
$
$
\begin{aligned}
& \therefore \quad I_2=\frac{13}{3} \cdot \int \frac{1}{3 x^2+4 x+5} \cdot d x \\
& =\frac{13}{3} \cdot \frac{1}{3} \cdot \int \frac{1}{x^2+\frac{4}{3} x+\frac{5}{3}} \cdot d x \\
& \because\left\{\left(\frac{1}{2} \text { coefficient of } t\right)^2\right. \\
& \left.=\left(\frac{1}{2}\left(\frac{4}{3}\right)\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\right\} \\
& \mathrm{I}_2=\frac{13}{9} \cdot \int \frac{1}{x^2+\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}+\frac{5}{3}} \cdot d x \\
& =\frac{13}{9} \cdot \int \frac{1}{x^2+\frac{4}{3} x+\frac{4}{9}+\frac{11}{9}} \cdot d x \\
& =\frac{13}{9} \cdot \int \frac{1}{\left(x+\frac{2}{3}\right)^2+\left(\frac{\sqrt{11}}{3}\right)^2} \cdot d x \\
& \because \quad \int \frac{1}{X^2+A^2} \cdot d x=\frac{1}{A} \tan ^{-1}\left(\frac{X}{A}\right)+c \\
& =\frac{13}{9} \cdot \frac{1}{\frac{\sqrt{11}}{3}} \cdot \tan ^{-1}\left(\frac{x+\frac{2}{3}}{\frac{\sqrt{11}}{3}}\right)+c_1 \\
& I_2=\frac{13}{3 \sqrt{11}} \cdot \tan ^{-1}\left(\frac{3 x+2}{\sqrt{11}}\right)+c_2 \ldots \ldots \\
&
\end{aligned}
$
thus, from (i), (ii) and (iii)
$
\begin{aligned}
& \therefore \quad \int \frac{2 x-3}{3 x^2+4 x+5} \cdot d x \\
& =\frac{1}{3} \cdot \log \left(3 x^2+4 x+5\right)-\frac{13}{3 \sqrt{11}} \cdot \tan ^{-1}\left(\frac{3 x+2}{\sqrt{11}}\right)+c \\
& \quad\left(\because c_1+c_2=c\right)
\end{aligned}
$
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Question 145 Marks
Evaluate : $\int \frac{\cos \theta}{\cos 3 \theta} \cdot d \theta$
Answer
$\begin{aligned}
& =\int \frac{\cos \theta}{4 \cos ^3 \theta-3 \cos \theta} \cdot d \theta \\
& =\int \frac{1}{4 \cos ^2 \theta-3} \cdot d \theta
\end{aligned}
$
Divide Numerator and Denominator by $\cos ^2 \theta$
$
\begin{aligned}
& I=\int \frac{\frac{1}{\cos ^2 \theta}}{\frac{4 \cos ^2 \theta-3}{\cos ^2 \theta}} \cdot d \theta \\
& =\int \frac{\sec ^2 \theta}{4-3 \sec ^2 \theta} \cdot d \theta \\
& =\int \frac{\sec ^2 \theta}{4-3\left(1+\tan ^2 \theta\right)} \cdot d \theta \\
& =\int \frac{\sec ^2 \theta}{1-3 \tan ^2 \theta} \cdot d \theta \\
& \text { put } \tan \theta=t \therefore \sec ^2 \theta \cdot d \theta=1 \cdot d t \\
& \mathrm{I}=\int \frac{1}{1-3 t^2} \cdot d t \\
& =\frac{1}{3} \cdot \int \frac{1}{\frac{1}{3}-t^2} \cdot d t \\
& =\frac{1}{3} \cdot \int \frac{1}{\left(\frac{1}{\sqrt{3}}\right)^2-t^2} \cdot d t \\
& =\frac{1}{3} \cdot \frac{1}{2\left(\frac{1}{\sqrt{3}}\right)} \cdot \log \left(\frac{\frac{1}{\sqrt{3}}+t}{\frac{1}{\sqrt{3}}-t}\right)+c \\
& =\frac{1}{2 \sqrt{3}} \cdot \log \left(\frac{1+\sqrt{3} t}{1-\sqrt{3} t}\right)+c \\
& =\frac{1}{2 \sqrt{3}} \cdot \log \left(\frac{1+\sqrt{3} \tan \theta}{1-\sqrt{3} \tan \theta}\right)+c \\
& \therefore \int \frac{\cos \theta}{\cos 3 \theta} \cdot d \theta=\frac{1}{2 \sqrt{3}} \cdot \log \left(\frac{1+\sqrt{3} \tan \theta}{1-\sqrt{3} \tan \theta}\right)+c \\
\end{aligned}
$
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Question 155 Marks
Evaluate : $\int \frac{1}{3+2 \sin ^2 x+5 \cos ^2 x} \cdot d x$
Answer
Divide Numerator and Denominator by $\cos ^2 x$
$
\begin{aligned}
& I=\int \frac{\frac{1}{\cos ^2 x}}{\frac{3+2 \sin ^2 x+5 \cos ^2 x}{\cos ^2 x}} \cdot d x \\
& =\int \frac{\sec ^2 x}{3 \sec ^2 x+2 \tan ^2 x+5} \cdot d x \\
& =\int \frac{\sec ^2 x}{3\left(1+\tan ^2 x\right)+2 \tan ^2 x+5} \cdot d x \\
& =\int \frac{\sec ^2 x}{5 \tan ^2 x+8} \cdot d x \\
& =\frac{1}{5} \cdot \int \frac{\sec ^2 x}{\tan ^2 x+\frac{8}{5}} \cdot d x \\
& \text { put } \tan x=t \therefore \sec ^2 x \cdot d x=1 \cdot d t \\
& \mathrm{I}=\frac{1}{5} \cdot \int \frac{1}{t^2+\frac{8}{5}} \cdot d t \\
& =\frac{1}{5} \cdot \int \frac{1}{t^2+\left(\frac{\sqrt{8}}{\sqrt{5}}\right)^2} \cdot d t \\
& =\frac{1}{5} \cdot \frac{1}{\frac{\sqrt{8}}{\sqrt{5}}} \cdot \tan ^{-1}\left(\frac{t}{\frac{\sqrt{8}}{\sqrt{5}}}\right)+c \\
& =\frac{1}{\sqrt{5}} \cdot \frac{1}{2 \sqrt{2}} \cdot \tan ^{-1}\left(\frac{\sqrt{5} t}{2 \sqrt{2}}\right)+c \\
& =\frac{1}{2 \sqrt{10}} \cdot \tan ^{-1}\left(\frac{\sqrt{5} \tan x}{2 \sqrt{2}}\right)+c \\
& \therefore \int \frac{1}{3+2 \sin ^2 x+5 \cos ^2 x} \cdot d x= \\
& \frac{1}{2 \sqrt{10}} \cdot \tan ^{-1}\left(\frac{\sqrt{5} \tan x}{2 \sqrt{2}}\right)+c \\
&
\end{aligned}
$
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Question 165 Marks
Evaluate : $\int \frac{1}{3-2 \sin x+5 \cos x} \cdot d x$
Answer
put $\tan \frac{x}{2}=t$
$
\begin{aligned}
& \therefore \quad d x=\frac{2}{1+t^2} \\
& \therefore \quad \sin x=\frac{2}{1+t^2} \cdot d t \text { and } \cos x=\frac{1-t^2}{1+t^2} \\
& I=\int \frac{1\left(\frac{2}{1+t^2}\right)}{3-2\left(\frac{2}{1+t^2}\right)+5\left(\frac{1-t^2}{1+t^2}\right)} \cdot d t \\
& =\int \frac{\frac{2}{1+t^2}}{\frac{3\left(1+t^2\right)-2(2 t)+5\left(1-t^2\right)}{1+t^2}} \cdot d t \\
& =\int \frac{2}{3+3 t^2-4 t+5-5 t^2} \cdot d t \\
& =\int \frac{2}{8-4 t-2 t^2} \cdot d t \\
& =\int \frac{1}{4-2 t-t^2} \cdot d t \\
& =\int \frac{1}{4-\left(t^2+2 t\right)} \cdot d t \\
& =\int \frac{1}{4-\left(t^2+2 t+1-1\right)} \cdot d t \\
& =\int \frac{1}{5-\left(t^2+2 t+1\right)} \cdot d t \\
& =\int \frac{1}{(\sqrt{5})^2-(t+1)^2} \cdot d t \\
& =\frac{1}{2(\sqrt{5})} \cdot \log \left(\frac{\sqrt{5}+(t+1)}{\sqrt{5}-(t+1)}\right)+c \\
& =\frac{1}{2 \sqrt{5}} \cdot \log \left(\frac{\sqrt{5}+1+\tan \frac{x}{2}}{\sqrt{5}-1-\tan \frac{x}{2}}\right)+c \\
&
\end{aligned}
$
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Question 175 Marks
Evaluate : $\int \frac{1}{2-3 \sin 2 x} \cdot d x$
Answer
put $\tan x=t$
$
\begin{aligned}
& \therefore \quad d x=\frac{1}{1+t^2} \cdot d t \text { and } \sin 2 x=\frac{2}{1+t^2} \\
& =\int \frac{1\left(\frac{1}{1+t^2}\right)}{2-3\left(\frac{2}{1+t^2}\right)} \cdot d t \\
& =\int \frac{\frac{1}{1+t^2}}{\frac{2\left(1+t^2\right)-3(2 t)}{1+t^2}} \cdot d t \\
& =\int \frac{1}{2+2 t^2-6 t} \cdot d t=\int \frac{1}{2\left(t^2-3 t+1\right)} \cdot d t \\
& \because\left\{\left(\frac{1}{2} \text { coefficient of } t\right)^2\right. \\
& \left.=\left(\frac{1}{2}(-3)\right)^2=\left(-\frac{3}{2}\right)^2=\frac{9}{4}\right\} \\
& =\frac{1}{2} \cdot \int \frac{1}{t^2-3 t+\frac{9}{4}-\frac{9}{4}+1} \cdot d t \\
& =\frac{1}{2} \cdot \int \frac{1}{\left(t^2-3 t+\frac{9}{4}\right)-\frac{5}{4}} \cdot d t \\
& =\frac{1}{2} \cdot \int \frac{1}{\left(t-\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2} \cdot d t \\
& =\frac{1}{2} \cdot \frac{1}{2\left(\frac{\sqrt{5}}{2}\right)} \cdot \log \left(\frac{\left(t-\frac{3}{2}\right)-\frac{\sqrt{5}}{2}}{\left(t-\frac{3}{2}\right)+\frac{\sqrt{5}}{2}}\right)+c \\
& =\frac{1}{2 \sqrt{5}} \cdot \log \left(\frac{2 t-3-\sqrt{5}}{2 t-3+\sqrt{5}}\right)+c \\
& =\frac{1}{2 \sqrt{5}} \cdot \log \left(\frac{2 \tan x-3-\sqrt{5}}{2 \tan x-3+\sqrt{5}}\right)+c \\
&
\end{aligned}
$
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Question 185 Marks
Evaluate : $\int(\sqrt{\tan x}+\sqrt{\cot x}) \cdot d x$
Answer
$
\begin{aligned}
& \text { I } \quad=\int\left(\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}\right) \cdot d x \\
& =\int \frac{\tan x+1}{\sqrt{\tan x}} \cdot d x \\
& \text { put } \quad \sqrt{\tan x}=t \quad \therefore \tan x=t^2 \therefore x=\tan ^{-1} t^2 \\
& \therefore \quad 1 \cdot d x=\frac{1}{1+\left(t^2\right)^2} \cdot 2 t \cdot d t \\
& \because \quad \sec ^2 x \cdot d x=2 t \cdot d t \\
& \therefore \quad d x=\frac{2 t}{\sec ^2 x} \cdot d x=\frac{2 t}{1+\tan ^2 x} \cdot d x=\frac{2 t}{1+t^4} \cdot d t \\
& =\int \frac{t^2+1}{t} \cdot \frac{2 t}{1+t^4} \cdot d t=2 \int \frac{t^2+1}{t^4+1} \cdot d t \\
& =2 \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} \cdot d t=2 \int \frac{\left(1+\frac{1}{t^2}\right)}{\left(t-\frac{1}{t}\right)^2+2} \cdot d t \\
& \text { put } \quad t-\frac{1}{t}=u \quad \because\left[\frac{d}{d t}\left(t-\frac{1}{t}\right)=1+\frac{1}{t^2}\right] \\
& \therefore\left(t-\left(-\frac{1}{t^2}\right)\right) d t=1 \cdot d u \\
& \therefore\left(1+\frac{1}{t^2}\right) d t=1 \cdot d u \\
& \mathrm{I}=2 \int \frac{1}{u^2+2} \cdot d u \\
& =2 \int \frac{1}{u^2+(\sqrt{2})^2} \cdot d u \\
& =2 \cdot \frac{1}{\sqrt{2}} \cdot \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+c \\
& =\sqrt{2} \cdot \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+c \\
& =\sqrt{2} \cdot \tan ^{-1}\left(\frac{t^2-1}{\sqrt{2} t}\right)+c \\
& =\sqrt{2} \cdot \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2} \cdot \sqrt{\tan x}}\right)+c \\
&
\end{aligned}
$
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Question 195 Marks
Evaluate : $\int \frac{\sin 2 x}{3 \sin ^4 x-4 \sin ^2 x+1} \cdot d x$
Answer
$\mathrm{I}=\int \frac{\sin 2 x}{3\left(\sin ^2 x\right)^2-4\left(\sin ^2 x\right)+1} \cdot d x$ put $\sin ^2 x=t$ $\therefore 2 \sin x \cdot \cos x \cdot d x=1 \cdot d t$ $\therefore \quad \sin 2 x \cdot d x=1 \cdot d t$
$=\int \frac{1}{3 t^2-4 t+1} \cdot d t$
$=\int \frac{1}{3\left(t^2-\frac{4}{3} t+\frac{1}{3}\right)} \cdot d t$
$\because\left\{\left(\frac{1}{2} \text { coefficient of } t\right)^2\right.$
$\left.=\left(\frac{1}{2}\left(-\frac{4}{3}\right)\right)^2=\left(-\frac{2}{3}\right)^2=\frac{4}{9}\right\}$
$I=\frac{1}{3} \cdot \int \frac{1}{t^2-\frac{4}{3} t+\frac{4}{9}-\frac{4}{9}+\frac{1}{3}} \cdot d t$
$=\frac{1}{3} \cdot \int \frac{1}{\left(t^2-\frac{4}{3} t+\frac{4}{9}\right)-\frac{1}{9}} \cdot d t$
$=\frac{1}{3} \cdot \int \frac{1}{\left(t-\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2} \cdot d t$
$=\frac{1}{3} \cdot \frac{1}{2\left(\frac{1}{3}\right)} \cdot \log \left(\frac{\left(t-\frac{2}{3}\right)-\frac{1}{3}}{\left(t-\frac{2}{3}\right)+\frac{1}{3}}\right)+c$
$=\frac{1}{2} \cdot \log \left(\frac{3 t-3}{3 t-1}\right)+c$
$=\frac{1}{2} \cdot \log \left(\frac{3 \sin ^2 x-3}{3 \sin ^2 x-1}\right)+c$
$\therefore \quad \int \frac{\sin 2 x}{3 \sin ^4 x-4 \sin ^2 x+1} \cdot d x$
$=\frac{1}{2} \cdot \log \left(\frac{3 \sin ^2 x-3}{3 \sin ^2 x-1}\right)+c$
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Question 205 Marks
Evaluate : $\quad \int \frac{1}{3-10 x-25 x^2} \cdot d x$
Answer
$
\begin{aligned}
& I=\int \frac{1}{25\left(\frac{3}{25}-\frac{10}{25} x-x^2\right)} \cdot d x \\
& =\int \frac{1}{25\left[\frac{3}{25}-\left(x^2+\frac{2}{5} x\right)\right]} \cdot d x \\
& \because\left\{\left(\frac{1}{2} \text { coefficient of } x\right)^2\right. \\
& \left.=\left(\frac{1}{2}\left(\frac{2}{5}\right)\right)^2=\left(\frac{1}{5}\right)^2=\frac{1}{25}\right\} \\
& =\frac{1}{25} \cdot \int \frac{1}{\frac{3}{25}-\left(x^2-\frac{2}{5} x+\frac{1}{25}-\frac{1}{25}\right)} \cdot d x \\
& =\frac{1}{25} \cdot \int \frac{1}{\frac{3}{25}-\left(x^2-\frac{2}{5} x+\frac{1}{25}\right)+\frac{1}{25}} \cdot d x \\
& =\frac{1}{25} \cdot \int \frac{1}{\frac{4}{25}-\left(x^2-\frac{2}{5} x+\frac{1}{25}\right)} \cdot d x \\
& =\frac{1}{25} \cdot \int \frac{1}{\left(\frac{2}{5}\right)^2-\left(x-\frac{1}{5}\right)^2} \cdot d x \\
& \because \quad \int \frac{1}{a^2-x^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{a+x}{a-x}\right)+c \\
& I=\frac{1}{25} \cdot \frac{1}{2\left(\frac{2}{5}\right)} \cdot \log \left(\frac{\frac{2}{5}+\left(x-\frac{1}{5}\right)}{\frac{2}{5}-\left(x-\frac{1}{5}\right)}\right)+c \\
& =\frac{1}{5} \cdot \log \left(\frac{1+5 x}{3-5 x}\right)+c \\
&
\end{aligned}
$
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Question 215 Marks
Prove The Following : $\int \frac{1}{\sqrt{x^2-a^2}} \cdot d x=\log \left(x+\sqrt{x^2-a^2}\right)+c$
Answer
Let $\quad \mathrm{I}=\int \frac{1}{\sqrt{x^2-a^2}} \cdot d x$
put $x=a \sec \theta \Rightarrow \quad \theta=\sec ^{-1}\left(\frac{x}{a}\right)$
$\therefore d x=a \cdot \sec \theta \cdot \tan \theta \cdot d \theta$
$I=\int \frac{1}{\sqrt{a^2 \sec ^2 \theta-a^2}} \cdot a \cdot \sec \theta \cdot \tan \theta \cdot d \theta$
$=\int \frac{a \cdot \sec \theta \cdot \tan \theta}{\sqrt{a^2\left(\sec ^2 \theta-1\right)}} \cdot d \theta$
$=\int \frac{a \cdot \sec \theta \cdot \tan \theta}{\sqrt{a^2 \cdot \tan ^2 \theta}} \cdot d \theta$
$=\int \frac{a \cdot \sec \theta \cdot \tan \theta}{a \cdot \tan \theta} \cdot d \theta$
$=\int \sec \theta \cdot d \theta$
$=\log (\sec \theta+\tan \theta)+c$
$=\log \left(\sec \theta+\sqrt{\sec ^2 \theta-1}\right)+c$
$=\log \left(\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right)+c_1$
$=\log \left(\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right)+c_1$
$=\log \left(\frac{x+\sqrt{x^2-a^2}}{a}\right)+c_1$
$=\log \left(x+\sqrt{x^2-a^2}\right)-\log a+c_1$
$=\log \left(x+\sqrt{x^2-a^2}\right)+c$
where $c=c_1-\log a$
$\therefore \quad \int \frac{1}{\sqrt{x^2-a^2}} \cdot d x=\log \left(x+\sqrt{x^2-a^2}\right)+c$
e.g. $\quad \int \frac{1}{\sqrt{x^2-16}} \cdot d x=\log \left(x+\sqrt{x^2-16}\right)+c$
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Question 225 Marks
Evaluate the following functions : $\int \frac{1}{1-\tan x} \cdot d x$
Answer
$
\begin{aligned}
& I=\int \frac{1}{1-\frac{\sin x}{\cos x}} \cdot d x \\
&=\int \frac{\cos x}{\cos x-\sin x} \\
&=\int \frac{\cos x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x-\frac{1}{\sqrt{2}} \cos x\right)} \cdot d x \\
&=\frac{1}{\sqrt{2}} \int \frac{\cos x}{\cos \frac{\pi}{4} \cos x-\sin \frac{\pi}{4} \sin x} \cdot d x \\
&=\frac{1}{\sqrt{2}} \int \frac{\cos x}{\cos \left(x+\frac{\pi}{4}\right)} \cdot d x \\
& \text { put } x+\frac{\pi}{4}=t \quad \therefore x=t-\frac{\pi}{4}
\end{aligned}
$
Differentiating both sides
$
1 \cdot d x=1 \cdot d t
$
$
\begin{aligned}
& =\frac{1}{\sqrt{2}} \int \frac{\cos \left(t-\frac{\pi}{4}\right)}{\cos t} \cdot d t \\
& =\frac{1}{\sqrt{2}} \int \frac{\cos t \cdot \cos \frac{\pi}{4}+\sin t \cdot \sin \frac{\pi}{4}}{\cos t} \cdot d t \\
& =\frac{1}{\sqrt{2}} \int\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \tan t\right] \cdot d t \\
& =\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}[t+\log (\sec t)]+c \\
& =\frac{1}{2}\left[x+\frac{\pi}{4}+\log \sec \left(x+\frac{\pi}{4}\right)\right]+c
\end{aligned}
$
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Question 395 Marks
Integrate the following w. r. t. x:

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

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