Mathematics

# $\displaystyle \int xsin^{2}xdx$

$\displaystyle \frac{x^{2}}{4}-\frac{xsin2x}{4}-\frac{1}{8}cos2x +c$

##### SOLUTION
$\int x\ sin^{2}x\ dx.$
$= x \int \ sin^{2}x.\ dx$ $-\int \left ( \int sin^{2}x.dx \right )dx$
$cos\ 2x= 1-2\ sin^{2}x$
$2\ sin^{2}x= \frac{1-cos\ 2x}{2}$
$x\ \int \left ( \frac{1- cos\ 2x}{2} \right )dx-\int \left ( \int \left ( \frac{1- cos\ 2x}{2} \right ).dx \right )dx.$
$= \frac{x}{2}[x- \ ^{1}/_{2}sin\ 2x]- \int [\frac{x}{2}- \ ^{1}/_{2}sin\ 2x]dx$
$= \frac{x^{2}}{2}- \frac{x}{4} sin\ 2x- \int [\frac{x^{2}}{4}+ \frac{1}{4} cos\ 2x]$
$= \frac{x^{2}}{4}- \frac{x\ sin\ 2x}{4}- \frac{1}{8} cos\ 2x+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 126

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