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

Question

1073 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-04T21:59:00

#= 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 :)