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