Is ?

Thanks.

Printable View

- February 2nd 2010, 08:29 AMnunos[SOLVED] Quick easy integration question
Is ?

Thanks. - February 2nd 2010, 08:30 AMHenryt999Integrate the product
Sure easy

Can you take it from here? - February 2nd 2010, 08:33 AMnunos
- February 2nd 2010, 08:36 AMe^(i*pi)
- February 2nd 2010, 08:38 AMHenryt999Sure

=

And you know how to integrate that? - February 2nd 2010, 08:46 AMnunos
What I thought was using what I believe to be some sort of the "power rule".

When one does

so, one can apply that here:

Is this thought process, correct, or have I made any mistake?

Thanks in advance. - February 2nd 2010, 09:37 AMHallsofIvy
Yes, that is correct. Let u= sin(x). Then du= cos(x) dx so your integral is

.

Note that you could also let u= cos(x). Then du= -sin(x)dx so your integral is

.

Since , . The two solutions are the same with C= C'- 1/2.

Henryt999 was pointing to a completely different but still valid solution. Since sin(2x)= 2sin(x)cos(x), sin(x)cos(x)= (1/2)sin(2x) and your integral becomes . Now let u= 2x so that du= 2 dx and dx= (1/2)du. Your integral is

.

But, now, so that solution is the same as . Replacing with , that becomes , the same as our second solution with and so the same as our first solution with . - February 2nd 2010, 09:45 AMnunos
Thanks HallsofIvy for your very clear explanation. I get it you know.