Find $\int \sin^2 x\, dx$

Question

Accepted Answer

$\frac{x}{2} - \frac{\sin 2x}{4} + C$. Using the identity sin²x = (1 - cos2x)/2, we get ∫sin²x dx = ∫(1 - cos2x)/2 dx = x/2 - sin2x/4 + C. Option B has an incorrect positive sign before sin2x, and option D uses cosine instead of sine.

Answer

A) $\frac{x}{2} - \frac{\sin 2x}{4} + C$

Answer

B) $\frac{x}{2} + \frac{\sin 2x}{4} + C$

Answer

C) $-\frac{\sin 2x}{2} + C$

Answer

D) $\frac{x}{2} - \frac{\cos 2x}{4} + C$

Find $\int \sin^2 x\, dx$

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