Integrate the function in Exercise: ^ 8 - ^ 8 x 1-2 ^ 2 x ^ 2 x - Vidyadip

Question

Integrate the function in Exercise:$\frac{\sin^{8}-\cos^{8}\text{x}}{1-2\sin^{2}\text{x}\cos^{2}\text{x}}$

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Answer

$\frac{\sin^{8}\text{x}-\cos^{8}\text{x}}{1-2\sin^{2}\text{x}\cos^{2}\text{x}}=\frac{\Big(\sin^{4}\text{x}+\cos^{4}\text{x)}(\sin^{4}\text{x}-\cos^{4}\text{x}\Big)}{\sin^{2}\text{x}+\cos^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}}$
$=\frac{\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)\Big(\sin^{2}\text{x}+\cos^{2}\text{x}\Big)\Big(\sin^{2}\text{x}-\cos^{2}\text{x}\Big)}{\Big(\sin^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}\Big)+\Big(\cos^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}\Big)}$
$=\frac{\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)\Big(\sin^{2}\text{x}-\cos^{2}\text{x}\Big)}{\Big(\sin^{2}\text{x}\Big(1-\cos^{2}\text{x}\Big)+\cos^{2}\text{x}\Big(1-\sin^{2}\text{x}\Big)}$
$=\frac{-\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)\Big(\cos^{2}\text{x}-\sin^{2}\text{x}\Big)}{\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)}$
$=-\cos2\text{x}$
$\therefore\int\frac{\sin^{8}\text{x}-\cos^{8}\text{x}}{1-2\sin^{2}\text{x}\cos^{2}\text{x}}\text{dx}=\int-\cos2\text{x}\ \text{dx}=-\frac{\sin2\text{x}}{2}+\text{C}$

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