First slide

Introduction to integration

Question

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

Easy

A

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

$- \frac{3}{4} \cos x + \frac{\cos 3 x}{12} + C$
C

$\frac{3}{4} \cos x - \frac{\cos 3 x}{12} + C$
D

None of these

Solution

$\begin{array}{l} \sin 3 x = 3 \sin x - 4 \sin^{3} x \\ \sin^{3} x = \frac{3 \sin x - \sin 3 x}{4} \\ \int \sin^{3} d x = \int \frac{3 \sin x - \sin 3 x}{4} d x \\ = \frac{3}{4} \int \sin x d x - \frac{1}{4} \int \sin 3 x d x \\ = \frac{3}{4} (- \cos x) - \frac{1}{4} (- \frac{\cos 3 x}{3}) + C \\ = - \frac{3}{4} \cos x + \frac{\cos 3 x}{12} + C \end{array}$

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