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

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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 :)

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