1 + ( x 4 ) , dx - Vidyadip

MCQ

$\int {\sqrt {1 + \sin \left( {\frac{x}{4}} \right)\,} } dx$

✓
$8\left( {\sin \frac{x}{8} - \cos \frac{x}{8}} \right) + c$
B
$\left( {\sin \frac{x}{8} + \cos \frac{x}{8}} \right) + c$
C
$\frac{1}{8}\left( {\sin \frac{x}{8} - \cos \frac{x}{8}} \right) + c$
D
$8\left( {\cos \frac{x}{8} - \sin \frac{x}{8}} \right) + c$

✓

Answer

Correct option: A.

$8\left( {\sin \frac{x}{8} - \cos \frac{x}{8}} \right) + c$

a
(a) $\int {\sqrt {1 + \sin \left( {\frac{x}{4}} \right)\,} } dx$
$ = \int {\sqrt {\left( {{{\sin }^2}\frac{x}{8} + {{\cos }^2}\frac{x}{8}} \right) + \left( {2\sin \frac{x}{8}\cos \frac{x}{8}} \right)} \,dx} $
$ = \int {\sqrt {{{\left( {\sin \frac{x}{8} + \cos \frac{x}{8}} \right)}^2}} dx} = \int {\left( {\sin \frac{x}{8} + \cos \frac{x}{8}} \right)} \,dx$
$ = \frac{{ - \cos \frac{x}{8}}}{{\left( {\frac{1}{8}} \right)}} + \frac{{\sin \frac{x}{8}}}{{\left( {\frac{1}{8}} \right)}} + c$ $ = 8\left( {\sin \frac{x}{8} - \cos \frac{x}{8}} \right) + c$.

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