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

1 Answer

  • 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
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