Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Evaluate the following: $\int\limits_0^{\Large \frac{\pi}{2}}\Large \frac{\tan xdx}{1+m^2\tan^2x}$

Can you answer this question?

1 Answer

0 votes
  • If $f(x)$ is substituted by $f(t)$,then $f'(x)dx=f'(t)dt$ Hence $\int f(x)dx= \int f(t)dt$
  • $ \int \large \frac{dx}{x}= \log |x|+c$
Step 1:
Let $I=\large\int\limits_0^{\Large \frac{\pi}{2}}\frac{\tan x}{1+m^2\tan^2x}$
we know $\tan x=\large\frac{\sin x}{\cos x}$
Hence $I=\int \limits_0^{\frac{\pi}{2}} \large \frac{\sin x / \cos x}{1+m^2. \Large\frac{\sin ^2x}{\cos ^2x}}dx$
On simplifying we get
$I=\large\int \limits_0^{\pi/2}\frac{\sin x.\cos x}{\cos ^2 x+m^2 \sin ^2 x}dx$
But $\sin x \cos x=\frac{1}{2} \sin 2x$
$I=\large\frac{1}{2}\int \limits_0^{\pi/2}\frac{\sin 2x}{\cos ^2 x+m^2 \sin ^2 x}dx$
Put $ m^2\sin ^2 x+\cos ^2 x=t$
On differentiating with respect to t we get,
$(m^2. 2\sin x \cos x+ 2 \cos x(-\sin x))dx=dt$
$=\sin 2x(m^2-1)dx=dt$
$ \sin 2xdx=\large\frac{dt}{m^2-1}$
Now substituting this in I, we get
$I=\large\frac{1}{2} \int \limits _{-1}^ {m^2} \frac{dt/m^2-1}{t}$
$=\large\frac{1}{2(m^2-1)} \int \limits_{-1}^{m^2} \frac{dt}{t}$
On integrating we get,
$I=\large\frac{1}{2 (m^2-1)} \bigg[\log|t|\bigg]_1^{m^2}+c$
Step 2:
Now applying the limits we get,
$I=\large\frac{1}{2 (m^2-1)} \bigg[\log m^2- \log (-1)\bigg]+c$
But $\log\; 1=0$
Therefore $I=\large\frac{1}{2 (m^2-1)} \log|m^2|+c$
$=\large\frac{1}{m^2-1} \log|m^2|$
$=\large\frac{1}{m^2-1} \log \sqrt {m^2}$
$=\large\frac{1}{m^2-1} (\log m)$
answered Apr 17, 2013 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App