Q. $\int_{0}^{\frac{\pi}{3}}\frac{\cos{x + \sin x}}{\sqrt{1 + \sin{2x}}}\ dx$ is equal to

A) $\frac{4\pi}{3}$
B) $\frac{2\pi}{3}$
C) $\pi$
D) $\frac{\pi}{3}$
Explanation

(d)
Let $I = \int_{0}^{\frac{\pi}{3}}\frac{\cos x + \sin x}{\sqrt{1 + 2\sin x\cos x}}dx$
$= \int_{0}^{\frac{\pi}{3}}\frac{\cos x + \sin x}{\sqrt{\cos^{2}x + \sin^{2}x + 2\sin{x\cos x}}}dx$
$= \int_{0}^{\frac{\pi}{3}}\frac{\cos x + \sin x}{\sqrt{\left( \cos x + \sin x \right)^{2}}}dx$
$= \int_{0}^{\frac{\pi}{3}}\frac{\cos x + \sin x}{\cos x + \sin x}dx$
$= \int_{0}^{\frac{\pi}{3}}1dx = \frac{\pi}{3}$

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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