Q. $\int_{}^{}{\sqrt{1 + \sin{\frac{x}{4}\ }}dx}$ is equal to

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

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

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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