How do you integrate #int sin^2alphacos^2alpha#?

Question

Charles Arocha · Accepted Answer

#= 1/8 int \ 1 - cos 4 alpha \ d alpha#

#int sin^2alphacos^2alpha \ d alpha#

#= int (sinalpha \ cos alpha)^2 \ d alpha#

use sine double-angle #sin 2 alpha = 2 sin alpha cos alpha# #= int (1/2 sin 2 alpha )^2 \ d alpha#

#= 1/4 int sin^2 2 alpha \ d alpha#

use cosine double-angle formula: #cos 2 A = 1 - 2 sin^2 A# with #A = 2 alpha#

#= 1/4 int (1 - cos 4 alpha)/2 \ d alpha#

It's trivial from there :)

Christopher Allers · Answer

undefined To integrate $ \int \sin^2(\alpha) \cos^2(\alpha) \, d\alpha $, you can use trigonometric identities:

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

Substitute these identities into the integral:

$$ \int \sin^2(\alpha) \cos^2(\alpha) \, d\alpha = \int \left(\frac{1 - \cos(2\alpha)}{2}\right) \left(\frac{1 + \cos(2\alpha)}{2}\right) \, d\alpha $$

$$ = \int \frac{1 - \cos^2(2\alpha)}{4} \, d\alpha $$

$$ = \frac{1}{4} \int (1 - \cos^2(2\alpha)) \, d\alpha $$

$$ = \frac{1}{4} \left(\int 1 \, d\alpha - \int \cos^2(2\alpha) \, d\alpha\right) $$

$$ = \frac{1}{4} \left(\alpha - \frac{1}{2}\int (1 + \cos(4\alpha)) \, d\alpha\right) $$

$$ = \frac{1}{4} \left(\alpha - \frac{1}{2}\left(\int 1 \, d\alpha + \int \cos(4\alpha) \, d\alpha\right)\right) $$

$$ = \frac{1}{4} \left(\alpha - \frac{1}{2}\left(\alpha + \frac{1}{4}\sin(4\alpha) \right)\right) + C $$

$$ = \frac{\alpha}{4} - \frac{\sin(4\alpha)}{8} + C $$

where $ C $ is the constant of integration.