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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integrals\[\int\limits_0^\frac{\pi}{2}\frac{\cos^2xdx}{\cos^2x+4\sin^2x}\]

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Toolbox:
  • (i)when f(x) is an integral function is substitute on t,then $f'(x)dx=dt.Hence\;\int f(x)dx=\int t.dt$
  • (ii)$\int \frac{dx}{a^2+x^2}=\frac{1}{a}\tan ^{-1}(x/a)+c$
Given $I=\int\limits_0^\frac{\pi}{2}\frac{\cos^2x}{\cos^2x+4\sin^2x}dx$
 
$\sin ^2 x=1-\cos ^2x$
 
Therefore $I=\int\limits_0^\frac{\pi}{2}\frac{\cos^2x}{\cos^2x+4(1-\cos^2x)}dx$
 
$=\int\limits_0^\frac{\pi}{2}\frac{\cos^2x}{\cos^2x+4-4\cos^2x}dx$
 
Multiply and divide 3 to the numerator and denominator
 
$=\frac{1}{3}\int\limits_0^\frac{\pi}{2}\frac{3cos^2x}{4-3\cos^2x}dx$
 
Add and subtract 4 in he numerator,
 
$I=\frac{1}{3}\int\limits_0^\frac{\pi}{2}\frac{4-3\cos^2x-4}{4-3\cos^2x}dx$
 
now seperating the terms
 
$I=\frac{1}{3}\int\limits_0^\frac{\pi}{2}\frac{4-3\cos^2x}{4-3\cos^2x}dx-\int\limits_0^\frac{\pi}{2}\frac{4}{4-3\cos^2x}dx$
 
$=\frac{1}{3}\int\limits_0^\frac{\pi}{2}dx-\int\limits_0^\frac{\pi}{2}\frac{4}{4-3\cos^2x}dx$
 
Consider $I_1=\int \frac{4}{4-3 \cos ^2x};$Multiply and divide each term by $\cos ^2x$
 
$I_1=\Large\int\frac{4/ \cos^2x}{4/\cos ^2x -\frac{3\cos^2x}{\cos^2x}}dx=\int \frac{4 \sec ^2x}{4\sec^2x-3}dx$
 
But $\sec^2x=1+\tan^2x$
 
Hence $I_1=\int \frac{4 \sec ^2x}{4(1+\tan^2x)-3}dx$
 
Hence $I=-\frac{1}{3}\int\limits_0^\frac{\pi}{2}dx+\frac{1}{3}\int\limits_0^\frac{\pi}{2}\frac{4 \sec ^2x}{1+4\tan^2x}dx$
 
Consider $I_1= \frac{1}{3}\int\limits_0^\frac{\pi}{2}\frac{4 \sec ^2x dx}{1+4\tan^2x}=\frac{2}{3}\int\limits_0^\frac{\pi}{2}\frac{2 \sec ^2x\;dx}{1+4\tan^2x}$
 
Put $t=2\tan x$; on differentiating w.r.t x,$ dt=2\sec^2xdx$
 
Limits also changes ;$when x=0;t=0 \;when\; x=\pi/2;t=\infty$
 
on substituting $\infty$
 
$I_1=\frac{2}{3}\int\limits_0^\infty\frac{dt}{1+t^2}$
 
$I=-\frac{1}{3}\int\limits_0^{\frac{\pi}{2}}dx+\frac{2}{3}\int\limits_0^\infty\frac{dt}{1+t^2}$
 
On integrating we get,
 
$I=-\frac{1}{3}[\pi/2-0+\frac{2}{3}[\tan^{-1}(t)]_0^\infty$
 
on applying limits we get
 
$I=-\frac{1}{3}[\pi/2-0]+\frac{2}{3}[\tan^{-1}(\infty)-\tan^{-1}(0)]$
 
$=-\frac{1}{3}(\pi/2)+\frac{2}{3}(\pi/2)$
 
$(tan^{-1}(\infty))=\pi/2$
 
Therefore $I=-\frac{\pi}{6}+\frac{2\pi}{6}=\frac{\pi}{6}$

 

 

answered Feb 19, 2013 by meena.p
 
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