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