First slide

Methods of integration

Question

$\int \sin^{3} {xcos}^{3} xdx$ is equal to

Moderate

A

$\frac{\sin^{6} x}{6} - \frac{\cos^{4} x}{4} + C$
B

$\frac{\sin^{6} x}{3} - \frac{\cos^{4} x}{2} + C$
C

$\frac{\cos^{6} x}{6} - \frac{\cos^{4} x}{4} + C$
D

None of these

Solution

$\int \sin^{3} x \cdot \cos^{3} xdx$
Let $\cos x = t \Rightarrow - \sin x = \frac{dt}{dx} \Rightarrow dx = - \frac{dt}{\sin x}$
$\begin{array}{l} ∴ \int \sin^{3} x \cdot \cos^{3} xdx = \int \sin^{3} x \cdot t^{3} \frac{dt}{- \sin x} \\ = - \int \sin^{2} x \cdot t^{3} dt = - \int t^{3} (1 - \cos^{2} x) dt \\ = - \int t^{3} (1 - t^{2}) dt = - \int (t^{3} - t^{5}) dt \\ = - [\frac{t^{4}}{4} - \frac{t^{6}}{6}] + C = \frac{\cos^{6} x}{6} - \frac{\cos^{4} x}{4} + C \end{array}$

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