How to integrate $int sin^2t cos^4t dt$ ?

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Answer 1 · 2015-10-16T17:28:11

We could write it as $int (cos^4x-cos^6x)dx$ then use power reduction formulas to integrate $cos^4x$ and $cos^6x$ separately.

It may be slightly simpler to rewrite as:

$int sin^2xcos^2xcos^2x dx = int (sinxcosx)^2cos^2x dx$

$= int(1/2(sin2x))^2 cos^2x dx$

Now expand the square in the first factor and use power reduction on $cos^2x$ to get

$1/8intsin^2 2x(1+cos 2x) dx$

$= 1/8intsin^2 2x dx +1/8 int sin^2 2xcos 2x dx$

$intsin^2 2x dx$ may be evaluated by power reduction and

$int sin^2 2xcos 2x dx$ is a $u = sin 2x$ substitution

How to integrate int sin^2t cos^4t dtint sin^2t cos^4t dt?