∫ ^ 8 x - ^ 8 x (1 - 2 ^ 2 x ^ 2 x) dx is equal to

∫ \dfrac{\sin^{8}x - \cos^{8}x}{

\left(1 - 2\sin^{2}x\cos^{2}x\right)}dx

is equal to

Options

A $- \dfrac{1}{2}\sin 2x + c$
B $- \sin^{2}x + c$
C $- \dfrac{1}{2}\sin x + c$
D $\dfrac{1}{2}\sin 2x + c$

Correct answer

A. $- \dfrac{1}{2}\sin 2x + c$

Step-by-step solution

$\sin^{8}x - \cos^{8}x = \left(\sin^{2}x - \cos^{2}x\right)\left(\sin^{2}x + \cos^{2}x\right)\left(\sin^{4}x + \cos^{4}x\right)$
and $\sin^{4}x + \cos^{4}x = {\left(\sin^{2}x + \cos^{2}x\right)}^{2} - 2\sin^{2}x\cos^{2}x = 1 - 2\sin^{2}x\cos^{2}x$
$∴ I = ∫ \dfrac{\sin^{8}x - \cos^{8}x}{\left(1 - 2\sin^{2}x\cos^{2}x\right)}dx$
$= ∫ \dfrac{\left(\sin^{2}x - \cos^{2}x\right)\left(\sin^{4}x + \cos^{4}x\right)}{\left(\sin^{4}x + \cos^{4}x\right)}dx$
$= - ∫ \cos 2xdx$

$= - \dfrac{1}{2}\sin 2x + c$ , where $c$ is the constant of integration.

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