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