## Calculus Simple Integration by Parts question

1. The problem is this:

The antiderivative from 0 to 1 of (e^x)*sin(Nx)dx

I tried integrating by parts several times and I'm just not sure if what I'm doing was correct. I keep hitting a dead end. I'm not sure if I'm supposed to IBP twice or substitute sin(Nx) for something else. Help please :) The rest of the questions rely on this answer so please and thank you for any assistance. O yea, and the answer has to have N in it. N is just a constant that can be replaced with any number. thanks again
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 This is a case where no matter what you choose to be your u and dv, the terms will never reach zero and it will seem that you have to do it forever. However, you can look at it from an algebraic standpoint. When you perform integration by parts twice, you should end up with the integral of what you started with. You can add only this integral part to the left hand side of your equation to get 2[(e^x)*sin(Nx)dx]. You then can evaluate whatever is left on the right side of the equation from 0 to 1 and divide by two. This will give you the value of your initial integral.
 $$Im \left( \int_0^1 e^x \cdot e^{iNx}\ \mbox{d}x \right)$$

## Calculus Simple Integration by Parts question

thank you very much for your quick response. i was on the right track but i just didnt see the algebraic part. calculus is like a giant puzzle! u have to look at these problems with an open mind. take care and have a nice day
-kyle