First slide

Methods of integration

Question

$\int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} d x is equal to$

Moderate

A

$\frac{1}{2} \sin 2 x + C$
B

$- \frac{1}{2} \sin 2 x + C$
C

$- \frac{1}{2} \sin x + C$
D

$- \sin^{2} x + C$

Solution

$\begin{aligned} \int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} d x = \int \frac{(\sin^{2} x - \cos^{2} x) (\sin^{2} x + \cos^{2} x) (\sin^{4} x + \cos^{4} x)}{1 - 2 \sin^{2} x \cos^{2} x} \end{aligned}$
= $\int \frac{(\sin^{2} x - \cos^{2} x) (1 - 2 \sin^{2} x \cos^{2} x)}{1 - 2 \sin^{2} x \cos^{2} x} d x = \int - \cos 2 x d x = - \sin 2 x + c$

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