Question

Integration!
\[\int \sin^2 x dx\]

Answer 1

use substituion to solve this one

Answer 2

I actually think I got this one, but I also think my method is sort of crazy.

\[u = \sin x\]\[du = \cos x\]\[dv = \sin x\]\[v = -\cos x\]
\[uv - \int v du\]\[-\sin x\cos x + \int \cos^2 x dx\]

\[\int \sin^2 x dx = -\sin x\cos x + \int 1 dx - \int \sin^2 x dx\]\[2\int \sin^2 x dx = -\sin x\cos x + \int 1 dx\]
So I get
\[\frac{x - \sin x\cos x}{2} + C \]
Which is apparently the right answer.

What is the "correct" way to do it?

Answer 3

Yeah that is definitely 1 way to do it...another would be to use the fact that

\[\large \sin^2xdx = \frac{1}{2}(1 - \cos(2x))\]

Integrate term by term

\[\int\limits_{}^{}\frac{1}{2}dx - \int\limits_{}^{}\frac{1}{2}\cos(2x)dx\]
\[\frac{1}{2}\int\limits_{}^{}dx - \frac{1}{2}\int\limits_{}^{}\cos(2x)dx\]
\[\frac{x}{2} - \frac{1}{2}\int\limits_{}^{}\cos(2x)dx\]

let u = 2x
du = 2dx
dx = du/2

\[\frac{x}{2} - \frac{1}{2}\int\limits_{}^{}\cos(u)\frac{du}{2}\]
\[\frac{x}{2} - \frac{1}{4}\int\limits_{}^{}\cos(u)du\]

\[\frac{x}{2} - \frac{1}{4}\sin (u)\]

replace u = 2x

\[\frac{x}{2} - \frac{1}{4}\sin (2x)\]

we know

\[\large \sin(2x) = 2\sin(x)\cos(x)\]

so

\[\frac{x}{2} - \frac{1}{2}\sin (x)cos(x)\]
\[\frac{x - sin(x)cox(x)}{2} + C\]

Answer 4

*typo in that last one*

\[\large \frac{x - \sin(x)\cos(x)}{2} + C\]

Answer 5

I'm not sure if that's actually any shorter!
But it might be a little more "sane." Maybe.

Thank you.

Answer 6

Yeah I know what you mean...it requires more memorization of identities.