Integrating Difficult Expression: \intsin{^2}(kx)dx

[raintrek]

I'm probably missing something obvious here, but I'm trying to integrate the following expression;

\intsin{^2}(kx)dx

I've tried doing it by part but with no luck. Is there some specific method I need to follow, or is it one of those I can only get by looking it up?

 

[malawi_glenn]

ok this is math, not physics.

Pesonally, I would use the Euler identity for sin(kx) and then things are straight forward, or use a trig-identity. But by heart I never rememeber so many of them, so I almost always use Euler.

p.s why not using TeX 100%? :)

 

[George Jones]

Let me add some detail to malawi_glenn's two useful ideas.

Either write \mathrm{sin}\left(kx\right) in terms of exponentials, or write \mathrm{sin}^2\left(kx\right) in terms of \mathrm{cos}\left(2kx\right).

 

[arildno]

Or, as a third method, use integration by parts+cyclicity of integral.

 

[pam]

\sin^2 x=[1-\cos(2x)]/2