I'm stuck finding the antiderivative to solve an integral. It's step 6 underlined with a "?".

I think it should be - ( 1/5 * e^5x * cos9x), then we add the C.

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- Oct 12th 2010, 12:35 PMsolidstatemathstuck finding antiderivative
I'm stuck finding the antiderivative to solve an integral. It's step 6 underlined with a "?".

I think it should be - ( 1/5 * e^5x * cos9x), then we add the C. - Oct 12th 2010, 12:49 PMAckbeet
You're not doing the integration by parts correctly. Your derivative of u is incorrect (you need the chain rule), and your integral of dv is incorrect (use a substitution).

- Oct 12th 2010, 12:55 PMpickslides
make $\displaystyle \displaystyle I = \int e^5x\cos 9x ~dx$

As as you have shown integrating by parts twice is the correct method.

Your derivatives need some attention.

$\displaystyle \displaystyle \int \cos 9x ~dx = \frac{1}{9}\sin 9x$

$\displaystyle \displaystyle \frac{d}{dx}( e^5x)=5e^5x$ - Oct 12th 2010, 01:43 PMsolidstatemath
Isn't d/dx of e^5x = e^4x * 5 ?

Also, I list it as simply du = e^5x dx because my teacher does that for some reason. For instance in the example of e^x he lists it as e^x = e^x dx. So I did the same.. I don't want to learn something the wrong way here I thought my way of showing it was correct.

So if I were to fix this would the rest be ok? - Oct 12th 2010, 01:50 PMpickslides