Q. $\int_{}^{}\frac{1}{\sin^{2}{x.\cos^{2}x}}\ dx$ is equal to
A) $\sin{x - \cos{x + c}}$
B) $\tan{x + \cot{x + c}}$
C) $\cos{x + \sin{x + c}}$
D) $\tan{x - \cot{x + c}}$
Explanation
(d)
$\int_{}^{}\frac{1}{\sin^{2}{x.\cos^{2}x}}\ dx = \int_{}^{}\frac{\sin^{2}x + \cos^{2}x}{\sin^{2}{x\cos^{2}x}}\ dx$
$= \int_{}^{}\left( \frac{1}{\cos^{2}x} + \frac{1}{\sin^{2}x} \right)dx$
$= \int_{}^{}{(\sec^{2}{x + cosec^{2}\ x)}}dx$
$= \tan x - \cot x + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.