_ ^ x 4 - ^2 x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\cos x\sqrt {4 - {{\sin }^2}x} } \;dx = $

A
$\frac{1}{2}\sin x\sqrt {4 - {{\sin }^2}x} - 2{\sin ^{ - 1}}\left( {\frac{1}{2}\sin x} \right) + c$
✓
$\frac{1}{2}\sin x\sqrt {4 - {{\sin }^2}x} + 2{\sin ^{ - 1}}\left( {\frac{1}{2}\sin x} \right) + c$
C
$\frac{1}{2}\sin x\sqrt {4 - {{\sin }^2}x} + {\sin ^{ - 1}}\left( {\frac{1}{2}\sin x} \right) + c$
D
None of these

✓

Answer

Correct option: B.

$\frac{1}{2}\sin x\sqrt {4 - {{\sin }^2}x} + 2{\sin ^{ - 1}}\left( {\frac{1}{2}\sin x} \right) + c$

b
(b) Putting $\sin x = t \Rightarrow \cos x\,dx = dt,$ we get
$\int_{}^{} {\cos x\sqrt {4 - {{\sin }^2}x\,} dx} = \int_{}^{} {\sqrt {4 - {t^2}} dt = \int_{}^{} {\sqrt {{{(2)}^2} - {t^2}} dt} } $
$ = \frac{t}{2}\sqrt {4 - {t^2}} + \frac{4}{2}{\sin ^{ - 1}}\frac{t}{2} + c$
$ = \frac{1}{2}\sin x\sqrt {4 - {{\sin }^2}x} + 2{\sin ^{ - 1}}\left( {\frac{1}{2}\sin x} \right) + c.$

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