How do you find the antiderivative of ${(sin x)}^{2}$ $(sinx)^2$ ?

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How do you find the antiderivative of ${(sin x)}^{2}$ $(sinx)^2$ ?

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Answer 1 · 2016-07-02T10:16:51

$= \frac{x}{2} - \frac{1}{4} sin 2 x + C$ $= x/2 -1/4 sin 2x + C$

Explanation:

us the identity $cos 2 A = 1 - 2 {sin}^{2} A \Rightarrow {sin}^{2} A = \frac{1}{2} (1 - cos 2 A)$ $cos 2A = 1 - 2 sin^2 A implies sin^2 A = 1/2(1-cos 2A)$

so

$int dx qquad sin^2 x$

$= 1/2 int dx qquad 1 - cos 2x$

$= 1/2 (x -1/2 sin 2x) + C$

$= x/2 -1/4 sin 2x + C$

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