Question

\(\displaystyle \int \sin^2x\,dx=\)

• A. \(\frac{x}{2}+\frac{\sin2x}{4}+c\)
• B. \(\frac{x}{2}-\frac{\cos2x}{4}+c\)
• C. \(\frac{x}{2}+\frac{\cos2x}{4}+c\)
• D. \(\frac{x}{2}-\frac{\sin2x}{4}+c\)

Hint:

Use the identity \(\sin^2x=\frac{1-\cos2x}{2}\) before integrating.

Answer 1

The Correct Option is D

We need to evaluate: \[ \int \sin^2x\,dx. \] Use the trigonometric identity: \[ \sin^2x=\frac{1-\cos2x}{2}. \] So, \[ \int \sin^2x\,dx = \int \frac{1-\cos2x}{2}\,dx. \] Separate the integral: \[ =\frac{1}{2}\int 1\,dx-\frac{1}{2}\int \cos2x\,dx. \] Now, \[ \int 1\,dx=x. \] Also, \[ \int \cos2x\,dx=\frac{\sin2x}{2}. \] Therefore, \[ \int \sin^2x\,dx = \frac{x}{2}-\frac{1}{2}\cdot\frac{\sin2x}{2}+c. \] \[ =\frac{x}{2}-\frac{\sin2x}{4}+c. \] Hence, \[ \int \sin^2x\,dx=\frac{x}{2}-\frac{\sin2x}{4}+c. \]