How to evaluate $\int\frac{ \sin^8 x - \cos^8x } { 1 - 2\sin^2 x \cdot \cos^2 x }\mathrm{d}x$?

Question

I would like to calculate the following integral: $$ \int\frac{ \sin^8 x - \cos^8x } { 1 - 2\sin^2 x \cdot \cos^2 x }\mathrm{d}x. $$

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I have tried to rewrite the integral as:

$$ \int\frac{( \sin^4 x + \cos^4x) \cdot ( \sin^2 x + \cos^2 x) \cdot ( \sin^2 x - \cos^2 x )} { 1 - 2\sin^2x \cdot \cos^2 x }\mathrm{d}x $$

but I don't know how to go on.

Answer 1

Hint: Your Integrand is equal to $$1-2\cos^2(x)$$ For the proof, Show that

$$(1-2\cos^2(x))(1-2\sin^2(x)\cos^2(x))-(\sin(x)^8-\cos(x)^8)=0$$ It is $$\sin^8(x)-\cos^8(x)=(\sin^2(x)-cos^2(x))(\sin^4(x)+cos^4(x))=$$ $$-\cos(2x)(\sin^4(x)+\cos^4(x))$$ and from

$$\sin^2(x)+\cos^2(x)=1$$ we get by squaring

$$\sin^4(x)+\cos^4(x)=1-2\sin^2(x)\cos^2(x)$$

Answer 2

Hint: $$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x.$$

Answer 3

Given $$\int \dfrac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}dx$$Now multiply numerator and denominator by $\dfrac{4\sec^4x}{\cos4x+3}$ and we get$$\int\dfrac{(\tan^2x-1)\sec^2x}{(1+\tan^2x)^2}$$ And now use substitution $u=\tan x$ and $du=\sec^2xdx$and if you find $$\int\dfrac{u^2-1}{(u^2+1)^2}du$$ you will get the answer