First slide

Methods of integration

Question

$\int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} d x =$

Moderate

A

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

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

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

$- \sin^{2} x$

Solution

$\begin{array}{l} \int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} {xcos}^{2} x} dx \\ \int \frac{(\sin^{4} x + \cos^{4} x) (\sin^{4} x - \cos^{4} x)}{1 - 2 \sin^{2} {xcos}^{2} x} d x \\ \int \frac{[{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x] (\sin^{2} x + \cos^{2} x) (\sin^{2} x - \cos^{2} x)}{1 - 2 \sin^{2} {xcos}^{2} x} \\ = \int \frac{(1 - 2 \sin^{2} {xcos}^{2} x) (- \cos 2 x)}{1 - 2 \sin^{2} {xcos}^{2} x} dx \\ = \int - \cos 2 xdx = - \frac{1}{2} \sin 2 x + C \end{array}$

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