Integrate (Sinx)^2

**Fredrik373** · May 22nd 2010, 12:26 AM

Hi!

How do I integrate (Sinx)^2 ?

**Prove It** · May 22nd 2010, 12:41 AM

Originally Posted by Fredrik373

Hi!

How do I integrate (Sinx)^2 ?

Recall that $\cos^2{x} + \sin^2{x} = 1$ and $\cos^2{x} - \sin^2{x} = \cos{2x}$ .

So if you subtract the second equation from the first...

$(\cos^2{x} + \sin^2{x}) - (\cos^2{x} - \sin^2{x}) = 1 - \cos{2x}$

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

$\sin^2{x} = \frac{1}{2} - \frac{1}{2}\cos{2x}$ .

Therefore:

$\int{\sin^2{x}\,dx} = \int{\left(\frac{1}{2} - \frac{1}{2}\cos{2x}\right)\,dx}$

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

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