Q. $\int_{}^{}{\frac{(\sin{\theta + \cos{\theta)}}}{\sqrt{\sin{2\theta}}}d\theta}$ is equal to
A) $\log{|\cos{\theta - \sin{\theta + \sqrt{\sin{2\theta}}}}}| + c$
B) $\log{|\sin{\theta - \cos{\theta + \sqrt{\sin{2\theta}}}}}| + c$
C) $\sin^{- 1}(\sin{\theta - \cos{\theta)}} + c$
D) $\sin^{- 1}(\sin{\theta + \cos{\theta)}} + c$
Explanation
(c)
Let $I = \int_{}^{}{\frac{\sin{\theta + \cos\theta}}{\sqrt{1 + \sin{2\theta - 1}}}d\theta}$
$= \int_{}^{}\frac{\sin{\theta + \cos\theta}}{\sqrt{1 - \left( \sin{\theta - \cos\theta} \right)^{2}}}d\theta$
Put $\sin{\theta - \cos\theta} = t$
$\Rightarrow \left( \cos{\theta + \sin\theta} \right)d\theta = dt$
$\therefore\ I = \int_{}^{}{\frac{1}{\sqrt{1 - t^{2}}}dt}$
$= \sin^{- 1}{t + c}$
$= \sin^{- 1}{\left( \sin{\theta - \cos\theta} \right) +}c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.