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