Evaluate $\int\limits_{0}^{\pi/2} \frac{\tan x}{1 + m^2 \tan^2 x} dx$

$\frac{1}{m^2 - 1} \log m$

The correct answer is Option (1) → $\frac{1}{m^2 - 1} \log m$

Let $I= \int\limits_{0}^{\pi/2} \frac{\tan x}{1 + m^2 \tan^2 x} dx$

$= \int\limits_{0}^{\pi/2} \frac{\frac{\sin x}{\cos x}}{1 + m^2 \frac{\sin^2 x}{\cos^2 x}} dx$

$= \int\limits_{0}^{\pi/2} \frac{\frac{\sin x}{\cos x}}{\frac{\cos^2 x + m^2 \sin^2 x}{\cos^2 x}} dx$

$= \int\limits_{0}^{\pi/2} \frac{\sin x \cos x}{1 - \sin^2 x + m^2 \sin^2 x} dx [∵\cos^2 x = 1 - \sin^2 x]$

$= \int\limits_{0}^{\pi/2} \frac{\sin x \cos x}{1 - \sin^2 x(1 - m^2)} dx$

Put $\sin^2 x = t$

$\Rightarrow 2 \sin x \cos x dx = dt$

For limit, when $x = 0$, then $\sin^2 0 = t \Rightarrow t = 0$

when $x = \frac{\pi}{2}$, then $\sin^2 \frac{\pi}{2} = t \Rightarrow t = 1$

$∴I = \frac{1}{2} \int\limits_{0}^{1} \frac{dt}{1 - t(1 - m^2)}$

$= \frac{1}{2} \left[ -\log |1 - t(1 - m^2)| \cdot \frac{1}{1 - m^2} \right]_0^1$

$= \frac{1}{2} \left[ -\log |1 - 1 + m^2| \cdot \frac{1}{1 - m^2} + \log |1| \cdot \frac{1}{1 - m^2} \right]$

$= \frac{1}{2} \left[ -\log |m^2| \cdot \frac{1}{1 - m^2} \right] = \frac{2}{2} \frac{\log m}{(m^2 - 1)}  [∵\log(1) = 0]$

$= \frac{1}{m^2 - 1} \log m [∵\log m^n = n \log m]$