Q. $\int_{}^{}\frac{\sin x\cos x}{\sqrt{1 - \sin^{4}x}}\ dx$ is equal to
A) $\frac{1}{2}\sin^{- 1}{(\sin^{2}x)} + c$
B) $\frac{1}{2}\cos^{- 1}{(\sin^{2}{x)}} + c$
C) $\tan^{- 1}{(\sin^{2}x)} + c$
D) $\tan^{- 1}{(2\sin^{2}x)} + c$
Explanation
(a)
Let $I = \int_{}^{}\frac{\sin x\cos x}{\sqrt{1 - \sin^{4}x}}\ dx$
Put $\sin^{2}x = t \Rightarrow 2\sin x\cos x\ dx = dt$
$\therefore I = \int_{}^{}\frac{dt}{2\sqrt{1 - t^{2}}} = \frac{1}{2}\sin^{- 1}t + c$
$= \frac{1}{2}\sin^{- 1}{(\sin^{2}x)} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.