First slide

Introduction to integration

Question

$\int \sin^{2} x \cos^{2} x d x =$

Easy

A

$\frac{1}{2} (x - \frac{\sin 4 x}{4}) + c$
B

$\frac{1}{8} (x - \frac{\sin 4 x}{4}) + c$
C

$\frac{1}{8} (x + \frac{\sin 4 x}{4}) + c$
D

$\frac{1}{2} (x + \frac{\sin 4 x}{4}) + c$

Solution

$\frac{1}{4} \int \sin^{2} (2 x) d x$ $= \frac{1}{4} \int \frac{1 - \cos 4 x}{2} d x$
$= \frac{1}{8} \int 1 d x - \frac{1}{8} \int \cos 4 x d x$

$= \frac{1}{8} [x - \frac{\sin 4 x}{4}] + c$

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