_ ^ x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\cos \sqrt x \;dx = } $

✓
$2[\sqrt x \sin \sqrt x + \cos \sqrt x ] + c$
B
$2[\sqrt x \sin \sqrt x - \cos \sqrt x ] + c$
C
$2[\cos \sqrt x - \sqrt x \sin \sqrt x ] + c$
D
$ - 2[\sqrt x \sin \sqrt x + \cos \sqrt x ] + c$

✓

Answer

Correct option: A.

$2[\sqrt x \sin \sqrt x + \cos \sqrt x ] + c$

a
(a) Put $\sqrt x = t \Rightarrow \frac{1}{{2\sqrt x }}\,dx = dt \Rightarrow dx = 2t\,dt,$ then it reduces to $\int_{}^{} {2t\,.\cos t\,dt} = 2\left[ {t\,.\,\sin t - \int_{}^{} {\sin t\,dt} } \right]$
$ = 2t\sin t + 2\cos t$$ = 2[\sqrt x \sin \sqrt x + \cos \sqrt x ] + c$.

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