Question

The value of $\underset{0}{\overset{\pi /2}{\int }}\frac{\tan x}{1+m^2{\tan }^{2}x}dx$ 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
$I=\underset{0}{\overset{\pi /2}{\int }}\frac{\tan x}{1+m^2{\tan }^{2}x}dx$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\sin x}{\cos x}}{1+m^2\frac{{\sin }^{2}x}{{\cos }^{2}x}}dx$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x\cos x}{{\cos }^{2}x+m^2{\sin }^{2}x}dx$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x\cos x}{1-{\sin }^{2}x+m^2{\sin }^{2}x}dx$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^2\cos x}{1-{\sin }^{2}x\left(1-m^2\right)}dx$
Put, ${\sin }^{2}x=t\Rightarrow 2\sin x\cos xdx=dt$
If $x=0,t=0$
If $x=\frac{\pi }{2}$, then $t=1$
$\therefore I=\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{dt}{1-t\left(1-m^2\right)}$
$=\frac{1}{2}{\left[-\log \left|1-t\left(1-m^2\right)\right|\times \frac{1}{1-m^2}\right]}_0^1$
$=\frac{1}{2}\left[\frac{1}{m^2-1}\left\{\log \left|1-\left(1-m^2\right)-\log (1)\right|\right\}\right]$
$=\frac{1}{2}\times \frac{1}{m^2-1}\log \left|m^2\right|=\frac{2\log m}{2\left(m^2-1\right)}=\frac{\log m}{m^2-1}$