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Maths Solve the Following Question.(3 Marks)

Indefinite Integration · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 49 questions.

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Solve the Following Question.(3 Marks)

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Question 13 Marks
Integrate the following with respect to the respective variable:

$\frac{\sin ^6 \theta+\cos ^6 \theta}{\sin ^2 \theta \cdot \cos ^2 \theta}$

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Question 63 Marks
Evaluate : $\int \frac{\left(x^2+1\right) \cdot e^x}{(x+1)^2} \cdot d x$
Answer
$
\begin{aligned}
\text { I } & =\int e^x\left[\frac{x^2+1}{(x+1)^2}\right] \cdot d x \\
& =\int e^x\left[\frac{x^2-1+2}{(x+1)^2}\right] \cdot d x \\
& =\int e^x\left[\frac{x^2-1}{(x+1)^2}+\frac{2}{(x+1)^2}\right] \cdot d x \\
& =\int e^x\left[\frac{x-1}{x+1}+\frac{2}{(x+1)^2}\right] \cdot d x
\end{aligned}
$
Here $f(x)=\frac{x-1}{x+1}$
$
\begin{aligned}
\Rightarrow & f^{\prime}(x)=\frac{(x+1)(1)-(x-1)(1)}{(x+1)^2}=\frac{2}{(x+1)^2} \\
& \because \quad \int\left[f(x)+f^{\prime}(x)\right] \cdot d x=e^x \cdot f(x)+c \\
\text { I } & =e^x \cdot\left(\frac{x-1}{x+1}\right)+c \\
\therefore & \int \frac{\left(x^2+1\right) \cdot e^x}{(x+1)^2} \cdot d x=e^x \cdot\left(\frac{x-1}{x+1}\right)+c
\end{aligned}
$
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Question 73 Marks
Evaluate : $\int e^{\tan ^{-1} x} \cdot\left(\frac{1+x+x^2}{1+x^2}\right) \cdot d x$
Answer
put $\tan ^{-1} x=t$
$
\therefore x=\tan t
$
differentiating $w . r . t . x$
$
\begin{aligned}
& \therefore \frac{1}{1+x^2} \cdot d x=1 \cdot d t \\
& \mathrm{I}=\int e^t \cdot\left[1+\tan t+\tan ^2 t\right] \cdot d t \\
& =\int e^t \cdot\left[\tan t+\left(1+\tan ^2 t\right)\right] \cdot d t \\
& =\int e^t \cdot\left[\tan t+\sec ^2 t\right] \cdot d t \\
& \text { Here } f(t)=\tan t \\
& \Rightarrow \quad f^{\prime}(t)=\sec ^2 t \\
& \mathrm{I}=e^t \cdot f(t)+c \\
& =e^t \cdot \tan t+c \\
& =e^{\tan ^{-1} x} \cdot x+c \\
& \therefore \int e^{\tan ^{-1} x} \cdot\left(\frac{1+x+x^2}{1+x^2}\right) \cdot d x=e^{\tan ^{-1} x} \cdot x+c \\
\end{aligned}
$
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Question 83 Marks
Evaluate : $\int x \cdot \sin ^{-1} x \cdot d x$
Answer
$\mathrm{I}=\int \sin ^{-1} x \cdot x \cdot d x \quad \ldots$. by LIATE
$
\begin{aligned}
& =\sin ^{-1} x \cdot \int x \cdot d x-\int \frac{d}{d x} \cdot \sin ^{-1} x \cdot \int x \cdot d x \cdot d x \\
& =\sin ^{-1} x \cdot \frac{x^2}{2}-\int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x-\frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x-\frac{1}{2} \int \frac{1-\left(1-x^2\right)}{\sqrt{1-x^2}} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x-\frac{1}{2} \int\left[\frac{1}{\sqrt{1-x^2}}-\frac{\left(1-x^2\right)}{\sqrt{1-x^2}}\right] \cdot d x \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x-\frac{1}{2} \int \frac{d x}{\sqrt{1-x^2}}+\frac{1}{2} \int \sqrt{1-x^2} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x-\frac{1}{2} \sin { }^{-1} x+\frac{1}{2}\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1}(x)\right]+c \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x+\frac{1}{4} x \sqrt{1-x^2}-\frac{1}{4} \sin ^{-1} x+c \\
\therefore \quad \int x \cdot \sin ^{-1} x \cdot d x & =\frac{1}{2} x^2 \cdot \sin ^{-1} x+\frac{1}{4} x \sqrt{1-x^2}-\frac{1}{4} \sin ^{-1} x+c
\end{aligned}
$
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Question 93 Marks
Evaluate : $\int \frac{x}{1-\sin x} \cdot d x$
Answer
$
\begin{aligned}
& \mathrm{I}=\int \frac{x}{1-\sin x} \cdot \frac{(1+\sin x)}{(1+\sin x)} \cdot d x \\
& =\int \frac{x(1+\sin x)}{1-\sin ^2 x} \cdot d x=\int \frac{x(1+\sin x)}{\cos ^2 x} \cdot d x=\int x \cdot\left(\frac{1}{\cos ^2 x}+\frac{\sin x}{\cos ^2 x}\right) \cdot d x \\
& =\int x \cdot\left(\sec ^2 x+\sec x \cdot \tan x\right) \cdot d x \\
& =\int x \cdot \sec ^2 x \cdot d x+\int x \cdot \sec x \cdot \tan x \cdot d x \\
& =\left(x \cdot \int \sec ^2 x \cdot d x-\int \frac{d}{d x} x \cdot \int \sec ^2 x \cdot d x \cdot d x\right)+\left(x \cdot \int \sec x \cdot \tan x \cdot d x-\int \frac{d}{d x} x \cdot \int \sec x \cdot \tan x \cdot d x \cdot d x\right) \\
& =x \cdot \tan x-\int(1) \cdot \tan x \cdot d x+x \cdot \sec x-\int(1) \cdot \sec x \cdot d x \\
& =x \cdot \tan x-\log (\sec x)+x \cdot \sec x-\log (\sec x+\tan x)+c \\
& =x \cdot(\sec x+\tan x)-\log (\sec x)-\log (\sec x+\tan x)+c \\
& \therefore \quad \int \frac{x}{1-\sin x} \cdot d x=x \cdot(\sec x+\tan x)-\log [(\sec x)(\sec x+\tan x)]+c \\
\end{aligned}
$
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Question 103 Marks
Evaluate : $\int x \cdot \tan ^{-1} x \cdot d x$
Answer
$
\begin{aligned}
\text { I } & =\int\left(\tan ^{-1} x \cdot\right) x \cdot d x \quad \ldots . \text { by LIATE } \\
& =\tan ^{-1} x \cdot \int x \cdot d x-\int \frac{d}{d x} \cdot \tan ^{-1} x \cdot \int x \cdot d x \cdot d x \\
& =\tan ^{-1} x \cdot \frac{x^2}{2}-\int \frac{1}{1+x^2} \cdot \frac{x^2}{2} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \tan ^{-1} x-\frac{1}{2} \int \frac{x^2}{1+x^2} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \tan ^{-1} x-\frac{1}{2} \int \frac{1+x^2-1}{1+x^2} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \tan ^{-1} x-\frac{1}{2} \int\left[1-\frac{1}{1+x^2}\right] \cdot d x \\
& =\frac{1}{2} x^2 \cdot \tan ^{-1} x-\frac{1}{2}\left[x-\tan ^{-1} x\right]+c \\
\therefore \quad \int x \cdot \tan ^{-1} x \cdot d x & =\frac{1}{2} x^2 \cdot \tan ^{-1} x-\frac{1}{2} x+\frac{1}{2} \tan ^{-1} x+c
\end{aligned}
$
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Question 113 Marks
Evaluate : $\int x^2 \cdot 5^x d x$
Answer
$
\begin{aligned}
\text {I} & =x^2 \cdot \int 5^x \cdot d x-\int \frac{d}{d x} \cdot x^2 \cdot \int 5^x \cdot d x \cdot d x \\
& =x^2 \cdot 5^x \cdot \frac{1}{\log 5}-\int 2 x \cdot 5^x \cdot \frac{1}{\log 5} \cdot d x \\
& =\frac{1}{\log 5} \cdot x^2 \cdot 5^x-\frac{2}{\log 5}\left\{x \cdot \int 5^x \cdot d x-\int \frac{d}{d x} \cdot x \cdot \int 5^x \cdot d x \cdot d x\right\} \\
& =\frac{1}{\log 5} \cdot x^2 \cdot 5^x-\frac{2}{\log 5}\left\{x \cdot 5^x \cdot \frac{1}{\log 5}-\int(1)\left(5^x \cdot \frac{1}{\log 5}\right) \cdot d x\right\} \\
& =\frac{1}{\log 5} \cdot x^2 \cdot 5^x-\frac{2}{\log 5}\left\{\frac{1}{\log 5} \cdot x \cdot 5^x \cdot \int \frac{1}{\log 5} \cdot 5^x \cdot d x\right\} \\
& =\frac{1}{\log 5} \cdot x^2 \cdot 5^x-\frac{2}{\log 5}\left\{\frac{1}{\log 5} \cdot x \cdot 5^x \cdot-\frac{1}{\log 5} \cdot 5^x \cdot \frac{1}{\log 5}\right\}+c \\
& =\frac{1}{\log 5} \cdot x^2 \cdot 5^x-\frac{2}{(\log 5)^2} \cdot x \cdot 5^x+\frac{2}{(\log 5)^3} \cdot 5^x+c \\
& =\frac{5^x}{\log 5} \cdot\left\{x^2-\frac{2 x}{\log 5}+\frac{2}{(\log 5)^2}\right\}+c
\end{aligned}
$
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Question 123 Marks
Evaluate : $\int \frac{1}{\sin x-\sqrt{3} \cos x} \cdot d x$
Answer
For any two positive numbers $a$ and $b$, we can find an angle $\theta$, such that
$
\therefore \quad \sin \theta=\frac{a}{\sqrt{a^2-b^2}} \text { and } \cos \theta=\frac{b}{\sqrt{a^2-b^2}}
$
Using this we express $\sin x-\sqrt{3} \cos x$
$
\begin{aligned}
& =\sqrt{1+3}(\cos \theta \cdot \sin x-\sin \theta \cdot \cos x) \\
& =2 \cdot \sin (x-\theta) \\
& =2 \cdot \sin \left(x-\frac{\pi}{3}\right) \\
\therefore I & =\int \frac{1}{2 \cdot \sin \left(x-\frac{\pi}{3}\right)} \cdot d x \\
& =\frac{1}{2} \cdot \int \operatorname{cosec}\left(x-\frac{\pi}{3}\right) \cdot d x \\
& =\frac{1}{2} \cdot \log \left(\operatorname{cosec}\left(x-\frac{\pi}{3}\right)-\cot \left(x-\frac{\pi}{3}\right)\right)+c \\
& =\frac{1}{2} \cdot \log \left(\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right)+c
\end{aligned}
$
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Question 133 Marks
Evaluate : $\int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \cdot d x$
Answer
$
\begin{aligned}
& \mathrm{I}=\int \frac{\sqrt{e^x}}{\sqrt{\frac{1}{e^x}-e^x}} \cdot d x \\
& =\int \frac{\sqrt{e^x}}{\sqrt{\frac{1-\left(e^x\right)^2}{e^x}}} \cdot d x \\
& =\int \frac{\sqrt{e^x}}{\frac{\sqrt{1-\left(e^x\right)^2}}{\sqrt{e^x}}} \cdot d x \\
& =\int \frac{\sqrt{e^x} \cdot \sqrt{e^x}}{\sqrt{1-\left(e^x\right)^2}} \cdot d x \\
& =\int \frac{e^x}{\sqrt{1-\left(e^x\right)^2}} \cdot d x \\
& \text { put } e^x=t \\
& \therefore \quad e^x \cdot d x=1 \cdot d t \\
& \mathrm{I}=\int \frac{1}{\sqrt{1-t^2}} \cdot d t \\
& =\sin ^{-1}(t)+c \\
& =\sin ^{-1}\left(e^x\right)+c \\
& \therefore \int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \cdot d x=\sin ^{-1}\left(e^x\right)+c \\
\end{aligned}
$
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Question 143 Marks
Evaluate : $\int \frac{1}{\sqrt{3 x^2-4 x+2}} \cdot d x$
Answer
$=\int \frac{1}{\sqrt{3\left(x^2-\frac{4}{3} x+\frac{2}{3}\right)}} \cdot d x$ $\because\left\{\left(\frac{1}{2} \text { coefficient of } x\right)^2=\left(\frac{1}{2}\left(-\frac{4}{3}\right)\right)^2=\left(-\frac{2}{3}\right)^2=\frac{4}{9}\right\}$
$\begin{aligned} & =\int \frac{1}{\sqrt{3} \cdot \sqrt{x^2-\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}+\frac{2}{3}}} \cdot d x \\ & =\frac{1}{\sqrt{3}} \cdot \int \frac{1}{\sqrt{\left(x^2-\frac{4}{3} x+\frac{4}{9}\right)+\left(\frac{2}{3}-\frac{4}{9}\right)}} \cdot d x \\ & =\frac{1}{\sqrt{3}} \cdot \int \frac{1}{\sqrt{\left(x-\frac{2}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2} \cdot d x}\end{aligned}$
$\begin{aligned} & \because \int \frac{1}{\sqrt{x^2+a^2}} \cdot d x=\log \left|x+\sqrt{x^2+a^2}\right|+c \\ & =\frac{1}{\sqrt{3}} \cdot \log \left(\left(x-\frac{2}{3}\right)+\sqrt{\left(x-\frac{2}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2}\right)+c \\ & =\frac{1}{\sqrt{3}} \cdot \log \left(\left(x-\frac{2}{3}\right)+\sqrt{x^2-\frac{4}{3} x+\frac{2}{3}}\right)+c\end{aligned}$
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Question 153 Marks
Evaluate the following functions : $\int \frac{e^x+1}{e^x-1} \cdot d x$
Answer
$
\begin{aligned}
& I=\int \frac{e^x-1+2}{e^x-1} \cdot d x \\
& =\int\left(\frac{e^x-1}{e^x-1}+\frac{2}{e^x-1}\right) \cdot d x \\
& =\int\left(1+\frac{2}{e^x-1}\right) \cdot d x \\
& =\int d x+\int \frac{2}{e^x\left(1-e^{-x}\right)} \cdot d x \\
& =\int 1 d x+2 \int \frac{e^{-x}}{1-e^{-x}} \cdot d x \\
& \text { put }\left(1-e^{-x}\right)=t \\
& \text { Differentiate w.r.t. } x \\
& -\left(e^{-x}\right)(-1) \cdot d x=1 d t \\
& e^{-x} \cdot d x=1 d t \\
& \mathrm{I}=\int 1 d x+2 \int \frac{1}{t} \cdot d t \\
& =x+2 \cdot \log (t)+c \\
& =x+2 \log \left(1-e^{-x}\right)+c \\
& \therefore \quad \int \frac{e^x+1}{e^x-1} \cdot d x=x+2 \log \left(1-e^{-x}\right)+c \\
\end{aligned}
$
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Question 193 Marks
Integrate the following functions w.r.t. x:

$e^{\sin ^{-1} x}\left[\frac{x+\sqrt{1-x^2}}{\sqrt{1-x^2}}\right]$

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Question 413 Marks
Integrate the following functions w.r.t. x:
$\frac{1}{x\left(x^3-1\right)}$
Answer
Let $I=\int \frac{1}{x\left(x^3-1\right)} d x$
$=\int \frac{x^{-4}}{x^{-4} x\left(x^3-1\right)} d x$
$=\int \frac{x^{-4}}{1-x^{-3}} d x$
$=\frac{1}{3} \int \frac{3 x^{-4}}{1-x^{-3}} d x$
$=\frac{1}{3} \int \frac{\frac{d}{d x}\left(1-x^{-3}\right)}{1-x^{-3}} d x$
$=\frac{1}{3} \log \left|1-x^{-3}\right|+c$
$\cdots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]$
$=\frac{1}{3} \log \left|1-\frac{1}{x^3}\right|+c$
$=\frac{1}{3} \log \left|\frac{x^3-1}{x^3}\right|+c .$
Alternative Method:
$\text { Let } I=\int \frac{1}{x\left(x^3-1\right)} d x$
$=\int \frac{x^2}{x^3\left(x^3-1\right)} d x$
Put $x^3=t \quad \therefore 3 x^2 d x=d t$
$\therefore x^2 d x=\frac{d t}{3}$
$\therefore I  =\int \frac{1}{t(t-1)} \cdot \frac{d t}{3}=\frac{1}{3} \int \frac{1}{t(t-1)} d t$
$ =\frac{1}{3} \int \frac{t-(t-1)}{t(t-1)} d t=\frac{1}{3} \int\left(\frac{1}{t-1}-\frac{1}{t}\right) d t$
$ =\frac{1}{3}\left[\int \frac{1}{t-1} d t-\int \frac{1}{t} d t\right]$
$=\frac{1}{3}[\log |t-1|-\log |t|]+c$
$=\frac{1}{3} \log \left|\frac{t-1}{t}\right|+c$
$=\frac{1}{3} \log \left|\frac{x^3-1}{x^3}\right|+c .$
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Question 423 Marks
Integrate the following functions w.r.t. x:
$\frac{7+4 x+5 x^2}{(2 x+3)^{\frac{3}{2}}}$
Answer
Let $I=\int \frac{7+4 x+5 x^2}{(2 x+3)^{\frac{3}{2}}} d x=\int \frac{5 x^2+4 x+7}{(2 x+3)^{\frac{3}{2}}} d x$
Put $2 x+3=t \quad \therefore 2 d x=d t$
$\therefore d x=\frac{d t}{2}$
Also, $x=\frac{t-3}{2}$
$\therefore I=\int \frac{5\left(\frac{t-3}{2}\right)^2+4\left(\frac{t-3}{2}\right)+7}{t^{\frac{3}{2}}} \cdot \frac{d t}{2}$
$=\frac{1}{2} \int \frac{5\left(\frac{t^2-6 t+9}{4}\right)+2(t-3)+7}{t^{\frac{3}{2}}} d t$
$=\frac{1}{2} \int \frac{5 t^2-30 t+45+8 t-24+28}{4 t^{\frac{3}{2}}} d t$
$=\frac{1}{8} \int \frac{5 t^2-22 t+49}{t^{\frac{3}{2}}} d t$
$=\frac{1}{8} \int\left(5 t^{\frac{1}{2}}-22 t^{-\frac{1}{2}}+49 t^{-\frac{3}{2}}\right) d t$
$=\frac{5}{8} \int t^{\frac{1}{2}} d t-\frac{22}{8} \int t^{-\frac{1}{2}} d t+\frac{49}{8} \int t^{-\frac{3}{2}} d t$
$=\frac{5}{8} \cdot \frac{t^{\frac{3}{2}}}{(3 / 2)}-\frac{11}{4} \cdot \frac{t^{\frac{1}{2}}}{(1 / 2)}+\frac{49}{8} \cdot \frac{t^{-\frac{1}{2}}}{(-1 / 2)}+c$
$=\frac{5}{12}(2 x+3)^{\frac{3}{2}}-\frac{11}{2} \sqrt{2 x+3}-\frac{49}{4} \cdot \frac{1}{\sqrt{2 x+3}}+c$
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