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Evaluate:$\int\limits_0^ \pi \large\frac{xdx}{a^2\cos^2x+b^2\sin^2x}$

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  • $(i)\; \int \limits_a^b f(x)dx=F(b)-F(a)$
  • $(ii)\; \int \limits_a^b f(x)dx=\int \limits_a^b f(a-x)dx$
Step 1:
Given $I=\int\limits_0^ \pi \large\frac{xdx}{a^2\cos^2x+b^2\sin^2x}$-----(1)
By applying the property $ \int \limits_a^b f(x)dx=\int \limits_a^b f(a-x)dx$
$I=\int\limits_0^ \pi \large\frac{\pi-x}{a^2\cos^2(\pi-x)+b^2\sin^2(\pi-x)}$
$I=\int\limits_0^ \pi \large\frac{\pi-x}{a^2\cos^2 x+b^2\sin^2 x}$$dx$-----(2)
because $\cos ^2 (\pi-x)=\cos ^2 x \qquad \sin ^2 (\pi-x)=\sin ^2 x$
Add equ(1) and equ(2)
$2I=\int\limits_0^ \pi \large\frac{x}{a^2\cos^2x+b^2\sin^2x}+\int\limits_0^ \pi \large\frac{\pi-x}{a^2\cos^2x+b^2\sin^2x}$$dx$
$2I=\int\limits_0^ \pi \large\frac{x+\pi-x}{a^2\cos^2x+b^2\sin^2x}$$dx$
$=\int\limits_0^ \pi \large\frac{\pi}{a^2\cos^2x+b^2\sin^2x}$$dx$
By applying the property
$\int \limits_0^{2a} f(x)dx=\int \limits_0^a f(2a-x) dx$ if $ (f(x)=f(2a-x)$
Clearly if $f(x)=\large\frac{1}{a^2\cos^2x+b^2\sin^2x}$ then
$f(2a-x)= \large\frac{1}{a^2\cos(\pi-x)+b^2\sin^2x}$
Step 2:
Hence $2I=2 \pi\int\limits_0^ {\pi/2} \large\frac{dx}{a^2\cos^2x+b^2\sin^2x}$
divide the numerator and denominator by $\cos ^2 x$
$2I=2 \pi \int \limits_0^{\pi/2} \large\frac{dx/\cos ^2 x}{\Large\frac{a^2 \cos^2 x}{\cos ^2 x}+\frac{b^2 \sin ^2 x}{\cos^2 x}}$$dx$
But $\large\frac{1}{\cos ^2 x}=$$\sec^2 x$
and $ \large\frac{\sin ^2 x}{\cos ^2 x}=$$\tan ^2 x$
$2I=2 \pi \int\limits_0^ {\pi/2} \large\frac{\sec^2 x}{a^2+b^2\tan^2x}$$dx$
Put $\tan x=t$ on differentiating w.r.t x $\sec ^2 dx=dt$
The limits also changes, when we substitute t,
when $x=0, \tan 0=0 \;therefore \;t=0$
when $x=\pi/2, \tan \pi/2=\infty \;therefore \;t=\infty$
Therefore $ 2I=2 \pi \int \limits_0^\infty \large\frac{dt}{a^2+b^2t^2}$
$ 2I=2 \pi \int \limits_0^\infty \large\frac{dt}{b^2\bigg(\frac{a^2}{b^2}+t^2\bigg)}$
$ 2I=\large\frac{2 \pi}{b^2} \int \limits_0^\infty \large\frac{dt}{\bigg(\frac{a}{b}\bigg)^2+t^2}$
This is of the form $\int\large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1}(x/a)$
Step 3:
Here $a=a/b\;and\;x=t$
On integrating we get
$2I=\large\frac{2\pi}{b^2} \bigg[\frac{b}{a} $$\tan ^{-1} \bigg(\large\frac{t}{a/b}\bigg)\bigg]$
$I=\large\frac{\pi}{b^2} \times \frac{b}{a} \bigg[$$\tan ^{-1} \bigg(\large\frac{bt}{a}\bigg)\bigg]_0^\infty$
On applying limits, $I=\large\frac{\pi}{ab} $$\bigg[\tan ^{-1} \infty-\tan 0\bigg]$
But $\tan ^{-1} \infty =\pi/2$ and $\tan 0=0$ Therefore $I=\large\frac{\pi}{ab} \times \frac{\pi}{2}=\frac{\pi^2}{2ab}$
Therefore $I=\large\frac{\pi^2}{2ab}$
answered Apr 25, 2013 by meena.p
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