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

Answer 1
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
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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