Question

Integrate: (looking for substitution or substitution by parts method, I think I \(\pm\) sign went wrong somewhere in my work...
\[\huge \int\limits_{0}^{\pi} \frac{\cos(8θ)}{e^{θ}} d\theta\]

Answer 1

It keeps getting uglier with each subsequent substitution so I'm guess I have something not canceling out or there's some trick I'm not seeing at this time. ;-)

Answer 2

do you know the\[\int e^x\sin x dx\]integral trick?
it's pretty well-known, and is the basis of how I would approach this problem

Answer 3

This is what I was getting for what you just wrote
u = e^x , du = e^x dx
dv = sin(x) dx , v = -cos(x)
\[\int e^x\sin x dx =-e^xcosx-\int e^x(-\cos x) dx\]

Answer 4

If it was just \(\int\)xsinx dx it would be... sin(x)-xcos(x)+C yes?

Answer 5

yes to you last post
and continue with the other if you can\[\int e^x\sin x dx =-e^xcosx-\int e^x(-\cos x) dx=?\]

Answer 6

Ah a second set of substitutions gives you \(e^x \sin x dx\) again... Hmm...

Anti-derivative of +cos(x) \(\rightarrow\) +sin(x), can move the - out of the integral as -(-1) = +1

We get the same thing twice, which if it works like any other variable we can double it so...
\[\frac{1}{2} e^xsin(x)-e^xcos(x)+C\]
Still good?

Answer 7

yes, but you missed the parentheses\[\frac12(e^2\sin x-e^x\cos x)+C\]so you are familiar with the "feedback" trick then?

Answer 8

(where we wind up with the same integral we started with, and use algebra to solve for it)

Answer 9

or maybe you missed the parentheses for lack of understanding something?
let me type it out so you can see...

Answer 10

now it should be pretty much the same story with your integral, with uglier numbers

Answer 11

\[\int e^x\sin x dx =-e^xcosx-\int e^x(-\cos x) dx\]\[\int e^x\sin x dx =-e^x\cos x+e^x\sin x-\int e^x\sin xdx\]adding\(\large\int e^x\sin xdx\) to both sides gives\[2\int e^x\sin xdx=e^x(\sin x-\cos x)+C\]which we can solve for the integral (note that C/2 is still just some unknown constant, so we're gonna still call it C again after division)\[\int e^x\sin xdx={e^x(\sin x-\cos x)\over2}+C\]

Answer 12

as I said earlier, same process, uglier numbers in your prob

Answer 13

Following what you've written (which is quite clear, ty)... and trying it with what I have to do (and yes you're right I forgot the parenthesis)

I have:
\[[\frac{8}{65} e^{-\theta } \sin(8 \theta )-\frac{1}{65} e^{-\theta} cos(8 \theta)+C]_0^{\pi}\]

Answer 14

Ty for your help again Turing :D

Answer 15

welcome :)