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Maths Solve the Following Question.(2 Marks)

Indefinite Integration · Maharashtra Board STD 12 Science (MSBSHSE - English Medium). 94 questions.

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Solve the Following Question.(2 Marks)

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Question 52 Marks
Evaluate the following : $\int \frac{2 x-7}{\sqrt{3 x-2}} d x$
Answer
Express $(2 x-7)$ in terms of $(3 x-2)$
$
\begin{aligned}
& 2 x-7=\frac{2}{3}(3 x-2)+\frac{4}{3}-7 \\
& =\frac{2}{3}(3 x-2)-\frac{17}{3} \\
& I=\int\left[\frac{\frac{2}{3}(3 x-2)-\frac{17}{3}}{\sqrt{3 x-2}}\right] \cdot d x \\
& =\int\left[\frac{\frac{2}{3}(3 x-2)}{\sqrt{3 x-2}}-\frac{\frac{17}{3}}{\sqrt{3 x-2}}\right] \cdot d x \\
& =\frac{2}{3} \int \sqrt{3 x-2} \cdot d x-\frac{17}{3} \int \frac{1}{\sqrt{3 x-2}} \cdot d x \\
& =\frac{2}{3} \int(3 x-2)^{\frac{1}{2}} \cdot d x-\frac{17}{3} \int \frac{1}{\sqrt{3 x-2}} \cdot d x \\
& =\frac{2}{3} \cdot \frac{(3 x-2)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)} \cdot \frac{1}{3}-\frac{17}{3} \cdot 2 \cdot(\sqrt{3 x-2}) \cdot \frac{1}{3}+c \\
& =\frac{4}{27} \cdot(3 x-2)^{\frac{3}{2}}-\frac{34}{9} \cdot(3 x-2)^{\frac{1}{2}}+c \\
&
\end{aligned}
$
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Question 62 Marks
Evaluate the following : $\int \frac{1}{\sqrt{3 x+1}-\sqrt{3 x-5}} \cdot d x$
Answer
$\int \frac{1}{\sqrt{3 x+1}-\sqrt{3 x-5}} \cdot d x$
$
\begin{aligned}
= & \int\left(\frac{1}{\sqrt{3 x+1}-\sqrt{3 x-5}}\right) \cdot\left(\frac{\sqrt{3 x+1}+\sqrt{3 x-5}}{\sqrt{3 x+1}+\sqrt{3 x-5}}\right) \cdot d x \\
& =\int \frac{\sqrt{3 x+1}+\sqrt{3 x-5}}{3 x+1-3 x+5} \cdot d x \\
& =\int \frac{\sqrt{3 x+1}+\sqrt{3 x-5}}{6} \cdot d x \\
& =\frac{1}{6} \cdot \int\left((3 x+1)^{\frac{1}{2}}+(3 x-5)^{\frac{1}{2}}\right) \cdot d x \\
& =\frac{1}{6} \cdot\left\{\int(3 x+1)^{\frac{1}{2}} \cdot d x+\int(3 x-5)^{\frac{1}{2}} \cdot d x\right\} \\
& =\frac{1}{6} \cdot\left\{\frac{(3 x+1)^{\frac{1}{2}}+1}{\left(\left(\frac{1}{2}+1\right) \cdot 3\right.}+\frac{(3 x-5)^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right) \cdot 3}\right\}+c \\
& =\frac{1}{18} \cdot\left\{\frac{2}{3}(3 x+1)^{\frac{3}{2}}+\frac{2}{3}(3 x-5)^{\frac{3}{2}}\right\}+c \\
& =\frac{1}{27} \cdot\left\{(3 x+1)^{\frac{3}{2}}+(3 x-5)^{\frac{3}{2}}\right\}+c
\end{aligned}
$
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Question 72 Marks
Evaluate : $\int e^x\left[\frac{x+2}{(x+3)^2}\right] \cdot d x$
Answer
$
\begin{aligned}
\mathrm{I} & =\int e^x\left[\frac{x+3-1}{(x+3)^2}\right] \cdot d x \\
& =\int e^x\left[\frac{x+3}{(x+3)^2}+\frac{-1}{(x+3)^2}\right] \cdot d x \\
& =\int e^x\left[\frac{1}{x+3}+\frac{-1}{(x+3)^2}\right] \cdot d x \\
& \therefore f(x)=\frac{1}{x+3} \Rightarrow f^{\prime}(x)=\frac{-1}{(x+3)^2} \\
& \therefore \int e^x\left[f(x)+f^{\prime}(x)\right] \cdot d x=e^x \cdot f(x)+c \\
& =e^x \cdot\left(\frac{1}{x+3}\right)+c \\
& =\frac{e^x}{x+3}+c \\
\therefore \quad & \int e^x\left[\frac{x+2}{(x+3)^2}\right] \cdot d x=\frac{e^x}{x+3}+c
\end{aligned}
$
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Question 82 Marks
Evaluate : $\int e^x\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) \cdot d x$
Answer
$
\begin{aligned}
\text { I } & =\int e^x\left(\frac{2+2 \sin x \cdot \cos x}{2 \cdot \cos ^2 x}\right) \cdot d x \\
& =\int e^x\left(\frac{1}{\cos ^2 x}+\frac{\sin x \cdot \cos x}{\cos ^2 x}\right) \cdot d x \\
& =\int e^x\left[\sec ^2 x+\tan x\right] \cdot d x \\
& =\int e^x\left[\tan x+\sec ^2 x\right] \cdot d x \\
& \therefore f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^2 x \\
& \therefore \int e^x\left[f(x)+f^{\prime}(x)\right] \cdot d x=e^x \cdot f(x)+c \\
& =e^{x \cdot \tan x+c} \\
\therefore & \int e^x\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) \cdot d x=e^x \cdot \tan x+c
\end{aligned}
$
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Question 102 Marks
Evaluate : $\int \frac{1}{\sqrt{3 x^2-7}} \cdot d x$
Answer
$
\begin{aligned}
& \text {I }=\int \frac{1}{\sqrt{3\left(x^2-\frac{7}{3}\right)} \cdot d x} \\
&=\int \frac{1}{\sqrt{3} \cdot \sqrt{x^2-\left(\frac{\sqrt{7}}{\sqrt{3}}\right)^2}} \cdot d x \\
&=\frac{1}{\sqrt{3}} \cdot \int \frac{1}{\sqrt{x^2-\left(\frac{\sqrt{7}}{\sqrt{3}}\right)^2}} \cdot d x \\
& \int \frac{1}{\sqrt{x^2-a^2}} \cdot d x=\log \left|x+\sqrt{x^2-a^2}\right|+c \\
& \mathrm{I}=\frac{1}{\sqrt{3}} \cdot \log \left(x+\sqrt{x^2-\left(\frac{\sqrt{7}}{\sqrt{3}}\right)^2}\right)+c \\
&=\frac{1}{\sqrt{3}} \cdot \log \left(x+\sqrt{x^2-\frac{7}{3}}\right)+c
\end{aligned}
$
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Question 112 Marks
Evaluate : $\int \frac{1}{a^2-b^2 x^2} \cdot d x$
Answer
\begin{aligned}
& \mathrm{I}=\int \frac{1}{b^2\left(\frac{a^2}{b^2}-x^2\right)} \cdot d x \\
& =\frac{1}{b^2} \cdot \int \frac{1}{\left(\frac{a}{b}\right)^2-x^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 \\
& \mathrm{I}=\frac{1}{b^2} \cdot \frac{1}{2\left(\frac{a}{b}\right)} \cdot \log \left(\frac{\frac{a}{b}+x}{\frac{a}{b}-x}\right)+c \\
& =\frac{1}{b^2} \cdot \frac{1}{2\left(\frac{a}{b}\right)} \cdot \log \left(\frac{\frac{a}{b}+x}{\frac{a}{b}-x}\right)+c \\
& =\frac{1}{2 a b} \cdot \log \left(\frac{a+b x}{a-b x}\right)+c \\
\end{aligned}
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Question 122 Marks
Evaluate : $\int \frac{1}{4 x^2+11} \cdot d x$
Answer
$
\begin{aligned}
& \text { Solution : I }=\int \frac{1}{4\left(x^2+\frac{11}{4}\right)} \cdot d x \\
& =\frac{1}{4} \cdot \int \frac{1}{x^2+\left(\frac{\sqrt{11}}{2}\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 \\
& \mathrm{I}=\frac{1}{4} \cdot\left[\frac{1}{\left(\frac{\sqrt{11}}{2}\right)}\right] \cdot \tan ^{-1}\left[\frac{x}{\left(\frac{\sqrt{11}}{2}\right)}\right]+c \\
& =\frac{1}{2 \sqrt{11}} \tan ^{-1}\left(\frac{2 x}{\sqrt{11}}\right)+c \\
\end{aligned}
$
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Question 132 Marks
Prove The Following : $\int \frac{1}{x \sqrt{x^2-a^2}} \cdot d x=\frac{1}{a} \sec ^{-1}\left(\frac{x}{a}\right)+c$
Answer
Let $\mathrm{I}=\int \frac{1}{x \sqrt{x^2-a^2}} \cdot d x$ put $x=a \sec \theta \Rightarrow \quad \theta=\sec ^{-1}\left(\frac{x}{a}\right)$
$
\begin{aligned}
\therefore \quad d x & =a \cdot \sec \theta \cdot \tan \theta \cdot d \theta \\
\mathrm{I} & =\int \frac{1}{a \sec \theta \sqrt{\ldots-a^2}} \cdot \ldots \ldots \\
& =\int \frac{\tan \theta}{\sqrt{a^2(\ldots \ldots)}} \cdot d \theta \\
& =\frac{1}{a} \int 1 \cdot d \theta \\
& =\frac{1}{a} \cdot \theta+c \\
& =\frac{1}{a} \cdot \sec ^{-1}\left(\frac{x}{a}\right)+c \\
\therefore \quad & \int \frac{1}{x \sqrt{x^2-a^2}} \cdot d x=\frac{1}{a} \sec ^{-1}\left(\frac{x}{a}\right)+c \\
\text { e.g. } \quad & \int \frac{1}{x \sqrt{x^2-64}} \cdot d x=\frac{1}{8} \sec ^{-1}\left(\frac{x}{8}\right)+c
\end{aligned}
$
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Question 142 Marks
Prove The Following : $\int \frac{1}{\sqrt{a^2-x^2}} \cdot d x=\sin ^{-1}\left(\frac{x}{a}\right)+c$
Answer
Let $I=\int \frac{1}{\sqrt{a^2-x^2}} \cdot d x$ put
$
\begin{gathered}
x=a \sin \theta \Rightarrow \quad \sin \theta=\frac{x}{a} \\
\therefore \quad \theta=\sin ^{-1}\left(\frac{x}{a}\right) \\
d x=a \cdot \cos \theta d \theta \\
\mathrm{I} \quad=\int \frac{1}{\sqrt{a^2-a^2 \sin ^2 \theta}} \cdot a \cdot \cos \theta d \theta \\
\mathrm{I} \quad=\int \frac{a \cdot \cos \theta}{a \sqrt{1-a^2 \sin ^2 \theta}} \cdot d \theta
\end{gathered}
$
$\begin{aligned} &=\int \frac{\cos \theta}{\cos \theta} \cdot d \theta \\ &=\int 1 \cdot d \theta \\ &=\theta+c \\ & \therefore \quad \int \frac{1}{\sqrt{a^2-x^2}} \cdot d x=\sin ^{-1}\left(\frac{x}{a}\right)+c \\ & \text { e.g. } \int \frac{1}{\sqrt{81-x^2}} \cdot d x=\sin ^{-1}\left(\frac{x}{9}\right)+c\end{aligned}$
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Question 152 Marks
Prove The Following : $\int \frac{1}{a^2-x^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{a+x}{a-x}\right)+c$
Answer
Consider,
$
\begin{aligned}
& \mathrm{I}=\int \frac{1}{a^2-x^2} \cdot d x \\
&=\int \frac{1}{(\ldots)(\ldots)} \cdot d x \\
&=\int \frac{1}{2 a} \cdot\left[\frac{1}{\ldots}-\frac{1}{\ldots}\right] \cdot d x \\
&=\frac{1}{2 a} \cdot \int\left[\frac{1}{\ldots}-\frac{1}{a+x}\right] \cdot d x \\
&=\frac{1}{2 a} \cdot[\log (a+x)-\log (a-x)]+c \\
&=\frac{1}{2 a} \cdot \log \left(\frac{\ldots}{\ldots}\right)+c \\
& \therefore \quad \int \frac{1}{a^2-x^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{a+x}{a-x}\right)+c \\
& \text { e.g. } \quad \int \frac{1}{16-x^2} \cdot d x=\frac{1}{2(4)} \log \left(\frac{4+x}{4-x}\right)+c
\end{aligned}
$
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Question 162 Marks
Prove The Following : $\int \frac{1}{x^2-a^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{x-a}{x+a}\right)+c$
Answer
$
\begin{aligned}
& \text { Let } \quad I=\int \frac{1}{x^2-a^2} \cdot d x \\
& =\int \frac{1}{(x+a)(x-a)} \cdot d x \\
& =\int \frac{1}{2 a} \cdot\left[\frac{1}{x-a}-\frac{1}{x+a}\right] \cdot d x \\
& =\frac{1}{2 a} \cdot \int\left[\frac{1}{x-a}-\frac{1}{x+a}\right] \cdot d x \\
& =\frac{1}{2 a} \cdot[\log (x-a)-\log (x+a)]+c \\
& =\frac{1}{2 a} \cdot \log \left(\frac{x-a}{x+a}\right)+c \\
& \therefore \quad \int \frac{1}{x^2-a^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{x-a}{x+a}\right)+c \\
& \text { e.g. } \int \frac{1}{x^2-9} \cdot d x=\frac{1}{2(3)} \log \left(\frac{x-3}{x+3}\right)+c \\
&
\end{aligned}
$
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Question 172 Marks
Evaluate the following functions : $\int \frac{\sin (x+a)}{\cos (x-b)} \cdot d x$
Answer
$
\begin{aligned}
& =\int \frac{\sin [(x-b)+(a+b)]}{\cos (x-b)} \cdot d x \\
& =\int \frac{\sin (x-b) \cdot \cos (a+b)+\cos (x-b) \cdot \sin (a+b)}{\cos (x-b)} \cdot d x \\
& =\int\left[\frac{\sin (x-b) \cdot \cos (a+b)}{\cos (x-b)}+\frac{\cos (x-b) \cdot \sin (a+b)}{\cos (x-b)}\right] \\
& =\int[\cos (a+b) \cdot \tan (x-b)+\sin (a+b)] \cdot d x \\
& =\cos (a+b) \cdot \log (\sec (x-b))+x \cdot \sin (a+b)+c
\end{aligned}
$
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Question 182 Marks
Evaluate the following functions : $\int(3 x+2) \sqrt{x-4} \cdot d x$
Answer
$
\begin{aligned}
& \text { put } \quad x-4=t \\
& \therefore \quad x=4+t
\end{aligned}
$
Differentiate
$
\begin{aligned}
& 1 \cdot d x=1 \cdot d t \\
= & \int[3(4+t)+2] \cdot \sqrt{t} \cdot d t \\
= & \int(14+3 t) \cdot t^{\frac{1}{2}} \cdot d t \\
= & \int\left(14 t^{\frac{1}{2}}+3 t^{\frac{3}{2}}\right) \cdot d t \\
= & 14 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+3 \frac{t^{\frac{5}{2}}}{\frac{5}{2}} \cdot d x \\
= & \frac{28}{3}(x-4)^{\frac{3}{2}}+\frac{6}{5}(x-4)^{\frac{5}{2}}+c
\end{aligned}
$
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Question 192 Marks
Evaluate the following functions : $\int \frac{1}{3 x+7 x^{-n}} \cdot d x$
Answer
$
\begin{aligned}
& \text { : Consider } \int \frac{1}{3 x+7 x^{-n}} \cdot d x \\
& \quad=\int \frac{1}{3 x+\frac{7}{x^n}} \cdot d x=\int \frac{1}{\frac{3 x^{n+1}+7}{x^n}} \cdot d x \\
& \quad=\int \frac{x^n}{3 x^{n+1}+7} \cdot d x \\
& \quad \text { put } 3 x^{n+1}+7=t \\
& \quad \text { Differentiate } w \cdot r \cdot t \cdot x \\
& \quad 3(n+1) x^n \cdot d x=d t \\
& \therefore \quad x^n \cdot d x=\frac{1}{3(n+1)} d t \\
& =\int \frac{1}{3(n+1)} \cdot d t \\
& =\frac{1}{3(n+1)} \cdot \log (t)+c \\
& =\frac{1}{3(n+1)} \cdot \log \left(3 x^{n+1}+7\right)+c
\end{aligned}
$
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Question 202 Marks
Evaluate the following : $\int \tan ^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}} \cdot d x$
Answer
$
\begin{aligned}
I & =\int \tan ^{-1} \sqrt{\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}} \cdot d x \\
& =\int \tan ^{-1} \sqrt{\frac{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}} \cdot d x \\
& =\int \tan ^{-1} \sqrt{\tan ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} \cdot d x \\
& =\int \tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right] \cdot d x \\
& =\int\left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot d x \\
& =\frac{\pi}{4} x-\frac{1}{2} \cdot \frac{x^2}{2}+c \\
& =\frac{\pi}{4} x-\frac{x^2}{4}+c
\end{aligned}
$
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Question 212 Marks
Evaluate the following : $\int \frac{\cos x-\cos 2 x}{1-\cos x} \cdot d x$
Answer
$
\begin{aligned}
& \int \frac{\cos x-\cos 2 x}{1-\cos x} \cdot d x \\
& =\int \frac{\cos x-(\ldots \ldots \ldots)}{1-\cos x} \cdot d x \\
& =\int \frac{\cos x-\ldots \ldots \ldots}{1-\cos x} \cdot d x \\
& =\int \frac{\cos x(1-\cos x)+\ldots \ldots \ldots}{1-\cos x} \cdot d x \\
& =\int\left[\cos x+\frac{\ldots \ldots \ldots}{1-\cos x}\right] \cdot d x \\
& =\int[\cos x+(1+\cos x)] \cdot d x \\
& =\int(1+2 \cos x) \cdot d x \\
& =x+2 \sin x+c
\end{aligned}
$
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Question 222 Marks
Evaluate the following : $\int\left(\frac{\cos x}{1-\cos x}\right) \cdot d x$
Answer
$
\begin{aligned}
I & =\int\left(\frac{\cos x}{1-\cos x}\right)\left(\frac{1+\cos x}{1+\cos x}\right) \cdot d x \\
& =\int \frac{\cos x(1+\cos x)}{1-\cos ^2 x} \cdot d x \\
& =\int\left(\frac{\left.\cos x+\cos ^2 x\right)}{\sin ^2 x}\right) \cdot d x \\
& =\int\left(\frac{\cos x}{\sin ^2 x}+\frac{\cos ^2 x}{\sin ^2 x}\right) \cdot d x \\
& =\int\left(\operatorname{cosec} x \cdot \cot x+\cot ^2 x\right) \cdot d x \\
& =\int\left(\operatorname{cosec} x \cdot \cot x+\operatorname{cosec}^2 x-1\right) \cdot d x \\
& =(-\operatorname{cosec} x)+(-\cot x)-x+c \\
& =-\operatorname{cosec} x-\cot x-x+c
\end{aligned}
$
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Question 232 Marks
Evaluate the following : $\int \frac{\sin ^3 x-\cos ^3 x}{\sin ^2 x \cdot \cos ^2 x} \cdot d x$
Answer
$\mathrm{I}=\int\left(\frac{\sin ^3 x}{\sin ^2 x \cdot \cos ^2 x}-\frac{\cos ^3 x}{\sin ^2 x \cdot \cos ^2 x}\right) \cdot d x$
$
=\int\left(\frac{\sin x}{\cos ^2 x}-\frac{\cos x}{\sin ^2 x}\right) \cdot d x
$
$\begin{aligned} & =\int\left(\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}-\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x}\right) \cdot d x \\ & =\int(\sec x \cdot \tan x-\operatorname{cosec} x \cdot \cot x) \cdot d x \\ & =\sec x-(-\operatorname{cosec} x)+c \\ I & =\sec x+\operatorname{cosec} x+c\end{aligned}$
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Question 242 Marks
Evaluate the following : $\int \sin 5 x \cdot \cos 7 x \cdot d x$
Answer
We know that
$
\begin{aligned}
& 2 \sin A \cdot \cos B=\sin (A+B)+\sin (A-B) \\
I & =\frac{1}{2} \int 2 \sin 5 x \cdot \cos 7 x \cdot d x \\
= & \frac{1}{2} \int[\sin (5 x+7 x)+\sin (5 x-7 x)] \cdot d x \\
= & \frac{1}{2} \int[\sin (12 x)+\sin (-2 x)] \cdot d x \\
= & \frac{1}{2} \int(\sin 12 x-\sin 2 x) \cdot d x \\
= & \frac{1}{2} \cdot\left[-\cos 12 x \cdot \frac{1}{12}+\cos 2 x \cdot \frac{1}{2}\right]+c \\
I= & -\frac{1}{24} \cos 12 x+\frac{1}{4} \cos 2 x+c
\end{aligned}
$
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Question 252 Marks
Evaluate the following : $\int \sin ^4 x \cdot d x$
Answer
$
\begin{aligned}
I & =\int\left(\sin ^2 x\right)^2 \cdot d x \\
& =\int\left(\frac{1}{2}(1-\cos 2 x)\right)^2 \cdot d x \\
& =\frac{1}{4} \cdot \int\left(1-2 \cos 2 x+\cos ^2 2 x\right) \cdot d x \\
& =\frac{1}{4} \cdot \int\left[1-2 \cos 2 x+\frac{1}{2}(1+\cos 4 x)\right] \cdot d x \\
& =\frac{1}{4} \cdot \int\left(1-2 \cos 2 x+\frac{1}{2}+\frac{1}{2} \cos 4 x\right) \cdot d x \\
& =\frac{1}{4} \cdot \int\left(\frac{3}{2}-2 \cos 2 x+\frac{1}{2} \cos 4 x\right) \cdot d x \\
& =\frac{1}{4} \cdot\left[\frac{3}{2} x-2 \sin 2 x \cdot \frac{1}{2}+\frac{1}{2} \sin 4 x \cdot \frac{1}{4}\right]+c \\
& =\frac{1}{4} \cdot\left[\frac{3}{2} x-\sin 2 x+\frac{1}{8} \sin 4 x\right]+c
\end{aligned}
$
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Question 262 Marks
Evaluate the following : $\int \sqrt{1+\sin 3 x} \cdot d x$
Answer
$
\begin{aligned}
I & =\int \sqrt{\cos ^2 \frac{3 x}{2}+\sin ^2 \frac{3 x}{2}+2 \sin \frac{3 x}{2} \cdot \cos \frac{3 x}{2}} \cdot d x \\
& =\int \sqrt{\left(\cos \frac{3 x}{2}+\sin \frac{3 x}{2}\right)^2 \cdot d x} \\
& =\int\left(\cos \frac{3 x}{2}+\sin \frac{3 x}{2}\right) \cdot d x \\
& =\sin \frac{3 x}{2} \cdot \frac{1}{\frac{3}{2}}-\cos \frac{3 x}{2} \cdot \frac{1}{\frac{3}{2}}+c \\
& =\frac{2}{3}\left(\sin \frac{3 x}{2}-\cos \frac{3 x}{2}\right)+c
\end{aligned}
$
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