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Maths Answer the following questions in short.

Indefinite Integration · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 59 questions.

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Answer the following questions in short.

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Question 51 Mark
Integrate the following w.r.t. x:

$\sqrt{x} \sec \left(x^{\frac{3}{2}}\right) \tan \left(x^{\frac{3}{2}}\right)$

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Question 81 Mark
Evaluate the following : $\int \frac{3^x-4^x}{5^x} \cdot d x$
Answer
$
\begin{aligned}
I & =\int\left(\frac{3^x}{5^x}-\frac{4^x}{5^x}\right) \cdot d x \\
& =\int\left[\left(\frac{3}{5}\right)^x-\left(\frac{4}{5}\right)^x\right] \cdot d x \\
& =\frac{\left(\frac{3}{5}\right)^x}{\log \frac{3}{5}}-\frac{\left(\frac{4}{5}\right)^x}{\log \frac{4}{5}}+c
\end{aligned}
$
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Question 91 Mark
Evaluate the following : $\int \frac{x^3}{x-1} \cdot d x$
Answer
$
\begin{aligned}
\text { I } & =\int \frac{x^3-1+1}{x-1} \cdot d x \\
& =\int\left(\frac{x^3-1}{x-1}+\frac{1}{x-1}\right) \cdot d x \\
& =\int\left(\frac{(x-1)\left(x^2+x+1\right)}{(x-1)}+\frac{1}{x-1}\right) \cdot d x \\
& =\int\left(x^2+x+1+\frac{1}{x-1}\right) \cdot d x \\
& =\frac{x^3}{3}+\frac{x^2}{2}+x+\log (x-1)+c
\end{aligned}
$
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Question 111 Mark
Evaluate the following : $\int \frac{e^{4 \log x}-e^{5 \log x}}{x^5} \cdot d x$
Answer
$\int \frac{e^{4 \log x}-e^{5 \log x}}{x^5} \cdot d x$
$
\begin{aligned}
& =\int \frac{e^{\log x^4}-e^{\log x^5}}{x^5} \cdot d x, \quad \because a^{\log _a f(x)}=f(x) \\
& =\int\left(\frac{x^4-x^5}{x^5}\right) \cdot d x \\
& =\int\left(\frac{1}{x}-1\right) \cdot d x \\
& =\log (x)-x+c
\end{aligned}
$
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Question 121 Mark
Evaluate the following : $\int \frac{\sqrt{x}+1}{x+\sqrt{x}} \cdot d x$
Answer
$
\begin{aligned}
& \int \frac{\sqrt{x}+1}{x+\sqrt{x}} \cdot d x \\
= & \int \frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)} \cdot d x \\
= & \int \frac{1}{\sqrt{x}} \cdot d x \\
= & 2 \cdot \int \frac{1}{2 \sqrt{x}} \cdot d x \\
= & 2 \sqrt{x}+c
\end{aligned}
$
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Question 131 Mark
Evaluate the following : $\int(\tan x+\cot x)^2 \cdot d x$
Answer
$\int(\tan x+\cot x)^2 \cdot d x$
$
\begin{aligned}
& =\int\left(\tan ^2 x+2 \tan x \cdot \cot x+\cot ^2 x\right) \cdot d x \\
& =\int\left(\tan ^2 x+2+\cot ^2 x\right) \cdot d x \\
& =\int\left(\sec ^2 x-1+2+\operatorname{cosec}^2 x-1\right) \cdot d x \\
& =\int\left(\sec ^2 x+\operatorname{cosec}^2 x\right) \cdot d x \\
& =\int \sec ^2 x \cdot d x+\int \operatorname{cosec}^2 x \cdot d x \\
& =\tan x+(-\cot x)+c \\
& =\tan x-\cot x+c
\end{aligned}
$
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Question 141 Mark
Evaluate the following : $\int\left(\sin x+\frac{1}{x}+\frac{1}{\sqrt[3]{x}}\right) \cdot d x$
Answer
$\int\left(\sin x+\frac{1}{x}+\frac{1}{\sqrt[3]{x}}\right) \cdot d x$
$
\begin{aligned}
& =\int \sin x \cdot d x+\int \frac{1}{x} \cdot d x+\int \frac{1}{\sqrt[3]{x}} \cdot d x \\
& =\int \sin x \cdot d x+\int \frac{1}{x} \cdot d x+\int x^{-\frac{1}{3}} \cdot d x \\
& =-\cos x+\log x+\frac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+c \\
& =-\cos x+\log x+\frac{x^{\frac{2}{3}}}{\frac{2}{3}}+c
\end{aligned}
$
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Question 151 Mark
Evaluate the following : $\int\left(x^3+3^x\right) \cdot d x$
Answer
$
\begin{aligned}
& \int\left(x^3+3^x\right) \cdot d x \\
= & \int x^3 \cdot d x+\int 3^x \cdot d x \\
= & \frac{x^4}{4}+\frac{3^x}{\log 3}+c
\end{aligned}
$
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Question 161 Mark
Evaluate : $\int \frac{1}{x^2+8 x+12} \cdot d x$
Answer
$
\begin{aligned}
\mathrm{I} & =\int \frac{1}{x^2+8 x+16-4} \cdot d x \\
& =\int \frac{1}{(x+4)^2-(2)^2} \cdot d x
\end{aligned}
$
$
\begin{array}{r}
\because \int \frac{1}{x^2-a^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{x-a}{x+a}\right)+c \\
\mathrm{I}=\frac{1}{2(2)} \cdot \log \left(\frac{(x+4)-2}{(x+4)+2}\right)+c \\
=\frac{1}{4} \cdot \log \left(\frac{x+2}{x+6}\right)+c \\
\therefore \quad \int \frac{1}{x^2+8 x+12} \cdot d x=\frac{1}{4} \cdot \log \left(\frac{x+2}{x+6}\right)+c
\end{array}
$
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Question 171 Mark
Evaluate the following functions : $\int \frac{e^x(1+x)}{\cos \left(x \cdot e^x\right)} \cdot d x$
Answer
put $x \cdot e^x=t$
Differentiating both sides
$
\begin{aligned}
& \left(x \cdot e^x+e^x \cdot 1\right) \cdot d x=1 d t \\
& e^x(1+x) \cdot d x=1 d t
\end{aligned}
$$\begin{aligned} \mathrm{I} & =\int \frac{1}{\cos t} \cdot d t \\ & =\int \sec t \cdot d t \\ & =\log (\sec t+\tan t)+c \\ & =\log \left(\sec \left(x e^x\right)+\tan \left(x e^r\right)\right)+c\end{aligned}$
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Question 181 Mark
Evaluate the following functions : $\int \frac{1}{1+e^{-x}} \cdot d x$
Answer
$I=\int \frac{1}{1+e^{-x}} \cdot d x$
$
\begin{aligned}
& =\int \frac{1}{1+\frac{1}{e^x}} \cdot d x \\
& =\int \frac{1}{\frac{e^x+1}{e^x}} \cdot d x \\
& =\int \frac{e^x}{e^x+1} \cdot d x \\
\because \quad & \frac{d}{d x}\left(e^x+1\right) \cdot d x=e^x \\
& =\log \left[e^x+1\right]+c
\end{aligned}
$
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Question 191 Mark
Evaluate the following functions : $\int \frac{1}{x+\sqrt{x}} \cdot d x$
Answer
$
\begin{aligned}
& \quad \mathrm{I}=\int \frac{1}{x+\sqrt{x}} \cdot d x \\
& =\int \frac{1}{\sqrt{x}(\sqrt{x}+1)} \cdot d x \\
& \text { put } \sqrt{x}+1=t \\
& \therefore \quad \frac{1}{2 \sqrt{x}} \cdot d x=1 \cdot d t \\
& \therefore \quad \frac{1}{\sqrt{x}} \cdot d x=2 \cdot d t \\
& =\int \frac{1}{t} \cdot 2 \cdot d t \\
& =2 \cdot \int \frac{1}{t} \cdot d t \\
& =2 \cdot \log (t)+c \\
& =2 \cdot \log (\sqrt{x}+1)+c
\end{aligned}
$
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Question 201 Mark
Evaluate the following functions : $\int \frac{\sec ^8 x}{\operatorname{cosec} x} \cdot d x$
Answer
$\begin{aligned}
\mathrm{I} & =\int \sec ^7 x \cdot \sec x \cdot \frac{1}{\operatorname{cosec} x} \cdot d x \\
& =\int \sec ^7 x \cdot \frac{1}{\cos x} \cdot \sin x \cdot d x \\
& =\int \sec ^7 x \cdot \tan x \cdot d x \\
& =\int \sec ^6 x \cdot \sec x \cdot \tan x \cdot d x \\
& \text { put } \sec x=t \\
& \therefore \quad \sec x \cdot \tan x \cdot d x=d t \\
& =\int t^6 \cdot d t \\
& =\frac{t^7}{7}+c \\
& =\frac{\sec ^7 x}{7}+c
\end{aligned}
$
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Question 211 Mark
Evaluate the following functions : $\int \frac{\cos \sqrt{x}}{\sqrt{x}} \cdot d x $
Answer
$\begin{aligned} &\text {Let } \mathrm{I}=\int \frac{\cos \sqrt{x}}{\sqrt{x}} \cdot d x \\ & \text { put } \sqrt{x}=t \\ & \therefore \frac{1}{2 \sqrt{x}} \cdot d x=1 \cdot d t \\ & \therefore \frac{1}{\sqrt{x}} \cdot d x=2 \cdot d t \\ & =2 \cdot \int \cos t \cdot d t \\ & =2 \cdot \sin t+c \\ & =2 \cdot \sin \sqrt{x}+c \\ & \end{aligned}$
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Question 221 Mark
Evaluate the following functions : $\int \frac{\cot (\log x)}{x} \cdot d x$
Answer
$
\begin{aligned}
& \text { Let } \mathrm{I}=\int \frac{\cot (\log x)}{x} \cdot d x \\
& \text { put } \log x=t \\
& \therefore \frac{1}{x} \cdot d x=1 \cdot d t \\
& =\int \cot t \cdot d t \\
& =\log (\sin t)+c \\
& =\log (\sin \log x)+c \\
&
\end{aligned}
$
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Question 231 Mark
Evaluate the following : $\int \tan ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right) \cdot d x$
Answer
$
\begin{aligned}
I & =\int \cot ^{-1}\left(\frac{1+\cos 2 x}{\sin 2 x}\right) \cdot d x \\
& =\int \cot ^{-1}\left(\frac{2 \cos ^2 x}{2 \sin x \cdot \cos x}\right) \cdot d x \\
& =\int \cot ^{-1}(\cot x) \cdot d x \\
& =\int x \cdot d x=\frac{x^2}{2}+c
\end{aligned}
$
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Question 241 Mark
Evaluate the following : $\int \sin ^{-1}(\cos 3 x) \cdot d x$
Answer
$
\begin{aligned}
\text { I } & =\int \sin ^{-1}\left(\sin \frac{\pi}{2}-3 x\right) \cdot d x \\
& =\int\left(\frac{\pi}{2}-3 x\right) \cdot d x \\
& =\frac{\pi}{2} x-3 \frac{x^2}{2}+c
\end{aligned}
$
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Question 251 Mark
Evaluate the following : $\int \frac{1}{1-\sin x} \cdot d x$
Answer
$
\begin{aligned}
\text { I } & =\int\left(\frac{1}{1-\sin x}\right)\left(\frac{1+\sin x}{1+\sin x}\right) \cdot d x \\
& =\int \frac{1+\sin x}{1-\sin ^2 x} \cdot d x \\
& =\int \frac{1+\sin x}{\cos ^2 x} \cdot d x \\
& =\int\left(\frac{1}{\cos ^2 x}+\frac{\sin x}{\cos ^2 x}\right) \cdot d x \\
& =\int\left(\sec ^2 x+\sec x \cdot \tan x\right) \cdot d x \\
& =\tan x+\sec x+c
\end{aligned}
$
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Question 261 Mark
Evaluate the following : $\int \cos ^3 x \cdot d x$
Answer
$\quad \cos 3 A=4 \cos ^3 A-3 \cos A$
$
\begin{aligned}
I & =\int \frac{1}{4}(\cos 3 x+3 \cos x) \cdot d x \\
& =\frac{1}{4}\left(\sin 3 x \cdot \frac{1}{3}+3 \cdot \sin x\right)+c \\
& =\frac{1}{12} \sin 3 x+\frac{3}{4} \sin x+c
\end{aligned}
$
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Question 381 Mark
Integrate the following functions w.r.t. x:

$\frac{x^{n-1}}{\sqrt{1+4 x^n}}$

Answer
Let $I=\int \frac{x^{n-1}}{\sqrt{1+4 x^n}} d x$
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Question 391 Mark
Integrate the following functions w.r.t. x:

$\frac{10^9+10^2 \cdot \log 10}{10^2+x^{10}}$

Answer
Let $I=\int \frac{10 x^9+10^x \cdot \log 10}{10^x+x^{10}} d x$

Put $10^x+x^{10}=t$

$\therefore\left(10^x \cdot \log 10+10 x^9\right) d x=d t$

$\therefore I=\int \frac{1}{t} d t=\log |t|+c$

$=\log \left|10^x+x^{10}\right|+c$

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Question 411 Mark
Integrate the following functions w.r.t. x:
$\sin ^4 x \cdot \cos ^3 x$
Answer
Let $I=\int \sin ^4 x \cdot \cos ^3 x d x$
$=\int \sin ^4 x \cdot \cos ^2 x \cdot \cos x d x$
$=\int \sin ^4 x\left(1-\sin ^2 x\right) \cos x d x$
Put $\sin x=t \quad \therefore \cos x d x=d t$
$\therefore I  =\int t^4\left(1-t^2\right) d t=\int\left(t^4-t^6\right) d t$
$ =\int t^4 d t-\int t^6 d t$
$ =\frac{t^5}{5}-\frac{t^7}{7}+c$
$=\frac{1}{5} \sin ^5 x-\frac{1}{7} \sin ^7 x+c$
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Question 421 Mark
Integrate the following functions w.r.t. x:
$\frac{e^{3 z}}{e^{3 x}+1}$
Answer
Let $I=\int \frac{e^{3 x}}{e^{3 x}+1} d x$
$\text { Put } e^{3 x}+1=t$
$\therefore 3 e^{3 x} d x=d t$
$\therefore e^{3 x} d x=\frac{d t}{3}$
$\therefore I=\int \frac{1}{t} \cdot \frac{d t}{3}=\frac{1}{3} \int \frac{1}{t} d t$
$=\frac{1}{3} \log |t|+c=\frac{1}{3} \log \left|e^{3 x}+1\right|+c .$
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Question 431 Mark
Integrate the following functions w.r.t. x:

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

Answer
Let $I =\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^2}} d x$

Put $\sin ^{-1} x=t . \quad \therefore \frac{1}{\sqrt{1-x^2}} d x=d t$

$\therefore I=\int t^{\frac{3}{2}} d t=\frac{t^{\frac{5}{2}}}{5 / 2}+c$

$=\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c$

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Question 441 Mark
Integrate the following functions w.r.t. x:

$\frac{(\log x)^n}{x}$

Answer
Let $I=\int \frac{(\log x)^n}{x} d x$

Put $\log x=t . \quad \therefore \frac{1}{x} d x=d t$

$\therefore I=\int t^n d t=\frac{t^{n+1}}{n+1}+c$

$=\frac{1}{n+1} \cdot(\log x)^{n+1}+c$

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Question 451 Mark
Evaluate:
$\int \sin 4 x \cos 3 x d x$
Answer
$\int \sin 4 x \cos 3 x d x$
$=\frac{1}{2} \int \sin 4 x \cos 3 x d x$
$=\frac{1}{2} \int[\sin (4 x+3 x)+\sin (4 x-3 x)] d x$
$=\frac{1}{2} \int \sin 7 x d x+\frac{1}{2} \int \sin x d x$
$=\frac{1}{2}\left(\frac{-\cos 7 x}{7}\right)-\frac{1}{2} \cos x+c$
$=-\frac{1}{14} \cos 7 x-\frac{1}{2} \cos x+c .$
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Question 461 Mark
Evaluate:

$\int \sqrt{1-\cos 2 x} \cdot d x$

Answer
$\begin{array}{c}\int \sqrt{1-\cos 2 x} d x \\ =\int \sqrt{2 \sin ^2 x} d x=\sqrt{2} \int \sin x d x\end{array}$

$=-\sqrt{2} \cos x+c$

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Question 471 Mark
Evaluate:

$\int \sqrt{1+\sin 2 x} \cdot d x$

Answer
$\begin{aligned} & \int \sqrt{1+\sin 2 x} d x \\ = & \int \sqrt{\cos ^2 x+\sin ^2 x+2 \sin x \cos x} d x \\ = & \int \sqrt{(\cos x+\sin x)^2} d x \\ = & \int(\cos x+\sin x) d x \\ = & \int \cos x d x+\int \sin x d x \\ = & \sin x-\cos x+c .\end{aligned}$
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Question 481 Mark
Evaluate:
$\int \frac{\tan x}{\sec x+\tan x} \cdot d x$
Answer
$ \int \frac{\tan x}{\sec x+\tan x} d x$
$=\int \frac{\tan x}{\sec x+\tan x} \times \frac{\sec x-\tan x}{\sec x-\tan x} d x$
$=\int \frac{\sec x \tan x-\tan ^2 x}{\sec ^2 x-\tan ^2 x} d x$
$=\int \frac{\sec x \tan x-\left(\sec ^2 x-1\right)}{1} d x$ $=\int \sec x \tan x d x-\int \sec ^2 x d x+\int 1 d x$
$=\sec x-\tan x+x+c$
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Question 491 Mark
Evaluate:
$\int \frac{\sin x}{1+\sin x} \cdot d x$
Answer
$ \int \frac{\sin x}{1+\sin x} d x$
$=  \int \frac{\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} d x$
$=  \int \frac{\sin x-\sin ^2 x}{1-\sin ^2 x} d x$
$= \int \frac{\sin x-\sin ^2 x}{\cos ^2 x} d x$
$=\int\left(\frac{\sin x}{\cos ^2 x}-\frac{\sin ^2 x}{\cos ^2 x}\right) d x$
$=\int\left(\frac{1}{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) d x-\int \tan ^2 x d x$
$=\int \sec x \tan x d x-\int\left(\sec ^2 x-1\right) d x$
$=\int \sec x \tan x d x-\int \sec ^2 x d x+\int 1 d x$
$=\sec x-\tan x+x+c .$
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Question 501 Mark
Evaluate:
$\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} \cdot d x$
Answer
$\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} d x$
$=\int \frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cdot \cos ^2 x} d x$
$=\int\left(\frac{1}{\sin ^2 x}-\frac{1}{\cos ^2 x}\right) d x$
$=\int \operatorname{cosec}^2 x d x-\int \sec ^2 x d x$
$=-\cot x-\tan x+c$
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