I can't find a way to solve this:

abitinomaq1

abitinomaq1

Answered question

2022-03-31

I can't find a way to solve this:
π2π8sin2(t)cos2(t)dt

Answer & Explanation

Ishaan Stout

Ishaan Stout

Beginner2022-04-01Added 14 answers

On the second quadrant π2<x<π, we have that
sinx>0, cosx<08sin2xcos2x
=8sinx(cosx)
=22sinxcosx
22π2πsinxcosxdx
=22π2πsinxcosxdx
=2212sin2xπ2π
=2(sinπsinπ2)
=2

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?