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MCQ

$\int_{}^{} {x{{\sin }^2}x\;dx = } $

A
$\frac{{{x^2}}}{4} + \frac{x}{4}\sin 2x + \frac{1}{8}\cos 2x + c$
B
$\frac{{{x^2}}}{4} - \frac{x}{4}\sin 2x + \frac{1}{8}\cos 2x + c$
C
$\frac{{{x^2}}}{4} + \frac{x}{4}\sin 2x - \frac{1}{8}\cos 2x + c$
✓
$\frac{{{x^2}}}{4} - \frac{x}{4}\sin 2x - \frac{1}{8}\cos 2x + c$

✓

Answer

Correct option: D.

$\frac{{{x^2}}}{4} - \frac{x}{4}\sin 2x - \frac{1}{8}\cos 2x + c$

d
(d)$\int_{}^{} {x{{\sin }^2}x\,dx} = \int_{}^{} {x\,.\,\frac{{(1 - \cos 2x)}}{2}\,dx} $
$ = \frac{1}{2}\left[ {\int_{}^{} {x\,dx} - \int_{}^{} {x\,.\,\cos 2x\,dx} } \right] = \frac{{{x^2}}}{4} - \frac{x}{4}\sin 2x - \frac{1}{8}\cos 2x + c$.

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