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Home  >>  CBSE XII  >>  Math  >>  Integrals
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By using the properties of definite integrals,evaluate the integral\[\int\limits_0^\frac{\pi}{2}\frac{\cos^5x\;dx}{\sin^5x+\cos^5x}\]

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Toolbox:
  • (i)$\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$
  • (ii)$\sin(\frac{\pi}{2}-x)=\cos x$
  • (iii)$ \cos (\frac{\pi}{2}-x)=\sin x$
Given $\int\limits_0^\frac{\pi}{2}\large\frac{\cos^5x\;}{\sin^5x+\cos^5x}dx----->(1)$
 
By applying property $\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$
 
$I=\int\limits_0^\frac{\pi}{2}\large\frac{\cos^5(\frac{\pi}{2}-x)\;}{\sin^5(\frac{\pi}{2}-x)+\cos^5(\frac{\pi}{2}-x)}$
 
But $ \sin(\frac{\pi}{2}-x)=\cos x$ and
 
$\cos(\frac{\pi}{2}-x)=\sin x$
 
Therefore $ I=\int\limits_0^\frac{\pi}{2}\large\frac{\sin^5x\;}{\cos^5x+\sin^5x}dx-----(2)$
 
Adding equ(1) and equ(2)
 
$ 2I=\int\limits_0^\frac{\pi}{2}\large\frac{\cos^5x}{\sin^5 x+\cos^5x}+\frac{\sin^5x}{\cos^5x+\sin^5x}dx$
 
$2I=\int \limits_0^{\frac{\pi}{2}}dx$
 
On integrating,
 
$2I=[x]_0^{\frac{\pi}{2}}$
 
On applying limits
 
$2I=\frac{\pi}{2}-0$
 
$ =\frac{\pi}{2}$
 
Therefore $ I=\frac{\pi}{4}$

 

answered Feb 13, 2013 by meena.p
 

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