Question

Evaluate: $\int {\sin }^{2}xdx$

• A. $\frac{x}{2}-\frac{\sin 2x}{4}+C$
• B. $\frac{x}{2}-\frac{\sin 2x}{2}+C$
• C. $\frac{x}{4}-\frac{\sin 2x}{4}+C$
• D. $\frac{x}{2}-\frac{\sin 2x}{3}+C$

Answer 1

The correct option is A $\frac{x}{2}-\frac{\sin 2x}{4}+C$

$\int {\sin }^{2}xdx=\int \frac{1-\cos 2x}{2}dx$ $(\because {\sin }^{2}x=\frac{1-\cos 2x}{2})$
$=\frac{1}{2}\int (1-\cos 2x)dx$
$=\frac{1}{2}\int dx-\frac{1}{2}\int \cos 2xdx$
$=\frac{x}{2}+(-\frac{1}{2})\frac{\sin 2x}{2}+C=\frac{x}{2}-\frac{\sin 2x}{4}+C$