Question

The value of $\int\limits_0^{\pi / 2} \frac{\tan x}{1+m^2 \tan ^2 x} d x$ is equal to

• A. $\frac{\log m}{m^2+1}$
• B. $\frac{\log m}{m^2-1}$
• C. $\frac{\log m}{m^2}$
• D. None of these

Answer 1

Answer: (B)

We have
$I =\int\limits_0^{\pi / 2} \frac{\tan x}{1+m^2 \tan ^2 x} d x $
$ =\int\limits_0^{\pi / 2} \frac{\frac{\sin x}{\cos x}}{1+m^2 \frac{\sin ^2 x}{\cos ^2 x}} d x $
$ =\int\limits_0^{\pi / 2} \frac{\sin x \cos x}{\cos ^2 x+m^2 \sin ^2 x} d x $
$=\int\limits_0^{\pi / 2} \frac{\sin x \cos x}{1-\sin ^2 x+m^2 \sin ^2 x} d x$
$ =\int\limits_0^{\pi / 2} \frac{\sin ^2 \cos x}{1-\sin ^2 x\left(1-m^2\right)} d x$
Put, $\sin ^2 x=t \Rightarrow 2 \sin x \cos x d x=d t$
If $x=0, t=0$
If $x=\frac{\pi}{2}$, then $t=1$
$\therefore I =\frac{1}{2} \int\limits_0^1 \frac{d t}{1-t\left(1-m^2\right)}$
$=\frac{1}{2}\left[-\log \left|1-t\left(1-m^2\right)\right| \times \frac{1}{1-m^2}\right]_0^1$
$ =\frac{1}{2}\left[\frac{1}{m^2-1}\left\{\log \left|1-\left(1-m^2\right)-\log (1)\right|\right\}\right]$
$ =\frac{1}{2} \times \frac{1}{m^2-1} \log \left|m^2\right|=\frac{2 \log m}{2\left(m^2-1\right)}=\frac{\log m}{m^2-1}$