How do you find the antiderivative of $\int {sin}^{2} x d x$ $int sin^2xdx$ ?

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Answer 1 · 2017-06-20T12:43:54

I got: $\frac{1}{2} [x - sin (x) cos (x)] + c$ $1/2[x-sin(x)cos(x)]+c$

I tried this:
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Answer 2 · 2017-06-20T15:55:12

We can start with the cosine double angle formula to derive another expression equivalent to ${sin}^{2} x$ $sin^2x$ :

$cos 2 x = {cos}^{2} x - {sin}^{2} x = 1 - 2 {sin}^{2} x$ $cos2x=cos^2x-sin^2x=1-2sin^2x$

Then:

$2 {sin}^{2} x = 1 - cos 2 x$ $2sin^2x=1-cos2x$

So:

$\int {sin}^{2} x d x = \frac{1}{2} \int (1 - cos 2 x) d x$ $intsin^2xdx=1/2int(1-cos2x)dx$

Integrating both of these term-by-term, with a substitution in the integration of $cos 2 x$ $cos2x$ gives:

$= \frac{1}{2} (x - \frac{1}{2} sin 2 x)$ $=1/2(x-1/2sin2x)$

Which can be rewritten using $sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$ :

$= \frac{1}{2} (x - sin x cos x) + C$ $=1/2(x-sinxcosx)+C$

How do you find the antiderivative of ∫sin2xdxint sin^2xdx?