_ ^ xx^2 ;dx is equal to - Vidyadip

MCQ

$\int_{}^{} {x\cos {x^2}\;dx} $ is equal to

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

✓

Answer

Correct option: D.

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

d
(d) Put ${x^2} = t \Rightarrow dt = 2x\,dx$
Given integral$ = \frac{1}{2}\int_{}^{} {\cos t\,dt} = \frac{1}{2}\sin t = \frac{1}{2}\sin {x^2} + c$.

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