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