Question

Evaluate \(\rm \int sin^2 x\;dx\)

1. \(\rm \frac{x}{2}+\frac{\sin 2x}{2} + c\)
2. \(\rm \frac{x}{2}+\frac{\sin 2x}{4} + c\)
3. \(\rm \frac{x}{2}-\frac{\sin 2x}{4} + c\)
4. \(\rm \frac{x}{2}+\frac{\cos 2x}{4} + c\)

Answer 1

Correct Answer - Option 3 : \(\rm \frac{x}{2}-\frac{\sin 2x}{4} + c\)

Concept:

1 + cos 2x = 2cos2 x

1 - cos 2x = 2sin2 x

\(\rm \int \cos x\;dx = \sin x + c\)

 

Calculation:

I = \(\rm \int sin^2 x\;dx\)

= \(\rm \int \frac{1-\cos 2x}{2}\;dx\)

= \(\rm \frac{1}{2}\int (1-\cos 2x)\;dx\)

= \(\rm \frac{1}{2} \left[x-\frac{\sin 2x}{2} \right ] + c\)

= \(\rm \frac{x}{2}-\frac{\sin 2x}{4} + c\)