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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