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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\int\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}\]

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Toolbox:
  • (i)$\sin ax=-\frac{1}{a}\cos ax+c$
  • (ii)$a^2-b^2=(a+b)(a-b)$
  • (iii)$\sin ^2x+\cos ^2x=1$
  • (iv)$\cos ^2x-\sin ^2x=\cos 2x$
Given $I=\large\int\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}dx$
 
Consider $\large\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}$
 
$\large\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}=\frac{(\sin ^4+\cos^4)(\sin ^4x-\cos ^4x)}{\sin^2x+\cos^2x-2\sin^2x \cos ^2x}$
 
$=\large\frac{(\sin ^4+\cos^4)(\sin^2x+\cos^2x(\sin^2x-\cos^2x)}{\sin^2x+\cos^2x-\sin^2x\cos ^2x-\sin^2x\cos ^2x}$
 
$=\large\frac{(\sin ^4+\cos^4)(1)(\sin^2x-\cos^2x)}{\sin^2x(1-\cos^2x)+\cos ^2x(1-\sin^2x)}$
 
But $(1-\cos^2x)=\sin^2x$
 
$(1-\sin^2x)=\cos^2x$
 
$\large\frac{(\sin ^4+\cos^4)(\sin^2x-\cos^2x)}{\sin^4x+\cos ^4x}$
 
$I=\int (\sin^2x-\cos ^2x)dx$
 
$I=-\int (\cos^2x-\sin ^2x)dx$
 
But $\cos ^2x-\sin^2x=\cos 2x$
 
Hence $I=-\int \cos 2x\;dx$
 
On integrating we get,
 
$I=-\frac{1}{2}\sin 2x+c$
 
$I=-\frac{1}{2}\sin2x+c$

 

 

answered Feb 16, 2013 by meena.p
 
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