# integration by part

• Sep 4th 2011, 07:13 AM
wintersoltice
integration by part
Find $\int e^{cos x}sin 2x$ $dx$

what i tried to do was

$\int e^{cos x}sin 2x dx$
$=\frac{-1}{sinx}e^{cosx}sin2x-\int \frac{-1}{sinx} e^{cosx}(2)(cos2x)$ $dx$
change $sin2x$ to $2sinxcosx$ and $cos 2x$ to $-2sin^2x+1$, then simplify.
$=-2e^{cosx}cosx+4\int (-sinx)(e^{cosx})$ $dx$ $-2\int \frac{1}{sinx}e^{cosx}$ $dx$
$=-2e^{cosx}cosx+4e^{cosx}-2[\frac{-1}{sinx}e^{cosx}(\frac{1}{sinx})-\int(\frac{-1}{sinx}e^{cosx}(-cosec x cotx)$ $dx]$

it's like never going to end.....
did i make a mistake somewhere?
or this approach is wrong?
• Sep 4th 2011, 07:27 AM
Jester
Re: integration by part
This what I might try. Your integral is

$2\int e^{\cos x} \sin x \cos x dx$
If you let $u = \cos x$ then $du = - \sin x\, dx.$

Therefore, your integral transforms to

$-2 \int u \,e^u\,du$.

Then integrate by parts.
• Sep 4th 2011, 07:36 AM
wintersoltice
Re: integration by part
Quote:

Originally Posted by Danny
This what I might try. Your integral is

$2\int e^{\cos x} \sin x \cos x dx$
If you let $u = \cos x$ then $du = - \sin x\, dx.$

Therefore, your integral transforms to

$-2 \int u \,e^u\,du$.

Then integrate by parts.

i see.
so, change sin2x to 2sin x cos x first before integrating.
then, how about the initial approach, will it work eventually?