First slide

Introduction to integration

Question

$\int \sin^{3} x d x =$

Easy

A

$\frac{1}{4} [\frac{\cos 3 x}{3} - 3 \cos x] + C$
B

$\frac{1}{4} [- \frac{\cos 3 x}{3} - 3 \cos x] + C$
C

$\frac{1}{4} [\frac{\cos 3 x}{3} + 3 \cos x] + C$
D

$\frac{1}{4} [- \frac{\cos 3 x}{3} + 3 \cos x] + C$

Solution

$\int \sin^{3} x d x$ $= \int [\frac{3}{4} \sin x - \frac{1}{4} \sin 3 x] d x$
$= \frac{- 3}{4} \cos x + \frac{1}{12} \cos 3 x + c$

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