Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 83 Marks
Question
Evaluate: $\int x \sin ^3 x \cos xdx$.
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Answer
We can write it as $\int x \sin ^2 x \sin x \cos x d x$
We also know that $2 \sin x \cdot \cos x=\sin 2 x$
$\int x \sin ^2 x \sin x \cos x d x$
$=\frac{1}{2} \int x \sin ^2 x \sin 2 x d x$
We also know that $\sin ^2 x=\frac{1-\cos 2 x}{2}$
$\frac{1}{2} \int x \sin ^2 x \sin 2 x d x=\frac{1}{2} \int x \cdot\left(\frac{1-\cos 2 x}{2}\right) \sin 2 x d x$
$=\frac{1}{2}\left[\left(\int \frac{x \sin 2 x}{2} d x-\int \frac{x \cos 2 x \sin 2 x}{2} d x\right)\right]$
Here $\operatorname{Sin} 4 x =2 \sin 2 x \cdot \cos 2 x$
$=\frac{1}{2}\left[\left(\int \frac{x \sin 2 x}{2} d x-\frac{1}{4} \int x \sin 4 x d x\right)\right]$
Using $\text{BY PART METHOD.}$
Here $x$ is first function and $\operatorname{Sin} 2 x$ and $\sin 4 x$ as the second function.
$\int \text { a.b.dx }=a \int b . dx-\int\left[\frac{d a}{d x} \cdot \int b d x\right] d x$
$=\frac{1}{2}\left[\left(\frac{1}{2}\left\{x \int \sin 2 xdx-\int\left(\frac{dx}{dx} \cdot \int \sin 2 xdx\right) dx\right\}\right)-\left(\frac{1}{4}\left\{x \int \sin 4 x-\int\left(\frac{dx}{dx} \cdot \int \sin 4 xdx\right) dx\right\}\right)\right]$
$=\frac{1}{2}\left[\left(\frac{1}{2}\left\{-x \frac{\cos 2 x}{2}+\int \frac{\cos 2 x}{2} d x\right\}\right)-\left(\frac{1}{4}\left\{-x \frac{\cos 4 x}{4}+\int \frac{\cos 4 x}{4} d x\right\}\right)\right]$
$=\frac{1}{2}\left[\left(\frac{1}{2}\left\{-x \frac{\cos 2 x}{2}+\frac{\sin 2 x}{4}\right\}\right)-\left(\frac{1}{4}\left\{-x \frac{\cos 4 x}{4}+\frac{\sin 4 x}{16}\right\}\right)\right]+c$
$=\frac{-x \cos 2 x}{8}+\frac{\sin 2 x}{16}+\frac{x \cos 4 x}{32}-\frac{\sin 4 x}{128}+c$
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