# integration by part

• Sep 4th 2011, 07:13 AM
wintersoltice
integration by part
Find $\displaystyle \int e^{cos x}sin 2x$$\displaystyle dx what i tried to do was \displaystyle \int e^{cos x}sin 2x dx \displaystyle =\frac{-1}{sinx}e^{cosx}sin2x-\int \frac{-1}{sinx} e^{cosx}(2)(cos2x) \displaystyle dx change \displaystyle sin2x to \displaystyle 2sinxcosx and \displaystyle cos 2x to \displaystyle -2sin^2x+1, then simplify. \displaystyle =-2e^{cosx}cosx+4\int (-sinx)(e^{cosx})$$\displaystyle dx$ $\displaystyle -2\int \frac{1}{sinx}e^{cosx} $$\displaystyle dx \displaystyle =-2e^{cosx}cosx+4e^{cosx}-2[\frac{-1}{sinx}e^{cosx}(\frac{1}{sinx})-\int(\frac{-1}{sinx}e^{cosx}(-cosec x cotx)$$\displaystyle 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

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

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

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

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

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