Question Detail

MTH 102 MTH 102 Past Questions Objective Question

Question

Obtain $\int \sin^2 x dx$.
Options
A $\frac{1}{2}x - \frac{\sin 2x}{2}$
B $\frac{1}{2}x + \frac{\sin 2x}{4}$
C $\frac{1}{2}x - \frac{\sin x}{4}$
D $\frac{1}{2}x - \frac{\sin 2x}{4}$
Correct Answer
Correct Answer

Option D is the correct answer.

Detailed Explanation

To integrate $\sin^2 x$, we use the trigonometric identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$.So, $\int \sin^2 x dx = \int \frac{1 - \cos(2x)}{2} dx$$= \int \left( \frac{1}{2} - \frac{1}{2} \cos(2x) \right) dx$$= \frac{1}{2}x - \frac{1}{2} \cdot \frac{\sin(2x)}{2} + C$$= \frac{1}{2}x - \frac{\sin(2x)}{4} + C$.Thus, option D is the correct answer.

Hint

Use the double-angle identity for $\sin^2 x$.

Question Info