How come integration of sin²2x cos 2x dx become [sin^3 (2x) ]/6?

Answer 1

Ezra Adams

Recall first, that #sin^2 2x = (sin2x)^2#.

Let #u = sin2x#,

then #du = cos(2x) *2 dx= 2cos(2x) dx#. #" "#(Chain rule)

So we get

#int sin^2 2x cos 2x dx = int(sin2x)^2 cos2x dx#

After substitution, we have

#1/2intu^2du = 1/6u^6 +C#

And reversing our substitution, we finish with

#1/6 (sin2x)^3 +C# which may also be written #(sin^3 2x)/6+C#

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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.

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