_ ^ x ^2 xdx = - Vidyadip

MCQ

$\int_{}^{} {x{{\cos }^2}} xdx = $

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

✓

Answer

Correct option: B.

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

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

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