First slide

Introduction to integration

Question

$I = \int \frac{\sin^{10} x - \cos^{8} x \sin^{2} x + \sin^{8} x \cos^{2} x - \cos^{10} x}{1 - 2 \sin^{2} x \cos^{2} x} d x$ is equal to

Easy

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

$\begin{array}{l} \sin^{10} x - \cos^{8} x \sin^{2} x + \sin^{8} x \cos^{2} x - \cos^{10} x \\ = \sin^{2} x (\sin^{8} x - \cos^{8} x) + \cos^{2} x (\sin^{8} x - \cos^{8} x) \\ = (\sin^{2} x + \cos^{2} x) (\sin^{8} x - \cos^{8} x) \\ = (\sin^{2} x - \cos^{2} x) (\sin^{2} x + \cos^{2} x) (\sin^{4} x + \cos^{4} x) \\ = (\sin^{2} x - \cos^{2} x) (1 - 2 \sin^{2} x \cos^{2} x) \\ = - \cos 2 x (1 - 2 \sin^{2} x \cos^{2} x) \\ I = - \int \cos 2 x d x = - \frac{\sin 2 x}{2} + C \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App