Mathematics

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


ANSWER

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


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
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