1. ## integral with e^2x

Hi,

I'm trying to solve the integral e^2xsin3x dx. I think I need to use the forumula int. usinudu=sinu-ucosu+c. Is that correct?

Would u be e^2x
du=e^2x(2)?????

So...int. usinudu=sinu-ucosu+c
=e^2xsine^2x=sine^2x-e^2xcose^2x+c

I don't think this is correct. Could someone please give me a hint?

Thank you

2. $\displaystyle I=\int e^{2x}\sin 3xdx=\int\left(\frac{e^{2x}}{2}\right)'\sin 3xdx=$
$\displaystyle =\frac{1}{2}e^{2x}\sin 3x-\frac{3}{2}\int e^{2x}\cos 3xdx=$
$\displaystyle =\frac{1}{2}e^{2x}\sin 3x-\frac{3}{2}\int\left(\frac{e^{2x}}{2}\right)'\cos 3xdx=$
$\displaystyle =\frac{1}{2}e^{2x}\sin 3x-\frac{3}{4}e^2x\cos 3x-\frac{9}{4}I$
Now, finish it.

4. Hi,

Now, finish it.
Can you please explain to me what the I's stand for and why you put '? Is the ' necessary? Can you explain what you did? I don't really understand it. We just started doing these types of problems.
Thank you very much

5. $I$ represents the initial integral.
I used integration by parts:
$\int f'(x)g(x)dx=f(x)g(x)-\int f(x)g'(x)dx$

6. See the link I gave you.

You can derive a general formula using integration by parts.

7. Can one do it with a forumla? We're supposed to be using formulas now.

Wouldn't int. usindu=sinu-ucosu+c work or do you have to do integration by parts?

Thank you

8. $\int x\sin\,dx$ & $\int e^x\sin x\,dx$ are two different integrals.

Use integration by parts.