I have worked an integral and reduced the

aanvarendbq28

aanvarendbq28

Answered question

2022-03-30

I have worked an integral and reduced the integral to
nπ+sin(nπ2)sin(3πn2)2nπ

Answer & Explanation

Alannah Farmer

Alannah Farmer

Beginner2022-03-31Added 11 answers

First notice that
sin(3nπ2)=sin(nπ2+nπ)
=sin(nπ2)cos(nπ)+cos(nπ2)sin(nπ)
=sin(nπ2)cos(nπ)+0
=(1)nsin(nπ2)
and hence
sin(nπ2)sin(3nπ2)=(1(1)n)sin(nπ2)
The above expression is 0 or 2 depending on n is even or odd. Now, you can find a bound easily
0(1(1)n)sin(nπ2)2
And finally
12nπ+sin(nπ2)sin(3πn2)2nπ12+1nπ

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?