Integration of Product of e, with variable exponent, and a trigonometric function

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July 25th 2012, 06:12 PM
Chuzzle

Integration of Product of e, with variable exponent, and a trigonometric function

I have been asked to find the definite integral, as described above, of a product of e, with a variable exponent, and a trigonometric function; an example would be the integral of e^x.cos(x). I have tried to use the integration by parts, or integration by reverse product rule differentiation (integral of udv/dx = uv - integral of vdu/dx), but have almost immediately realised that because both the differentiation and integration either of e with a variable exponent, or a trigonomnetric function, does not gradually cancel out the variables but increases the complexity of the function, the integral will simply increase to an infinite length. I therefore assumed this technique to be incorrect, and I want to ask if anyone can help me with the proper technique? Any help would be much appreciated!
July 25th 2012, 08:22 PM
Soroban

Re: Integration of Product of e, with variable exponent, and a trigonometric function

Hello, Chuzzle!

Quote:

I have been asked to find the definite integral, as described above, of a product of e, with a variable exponent,
and a trigonometric function. .An example would be: . $\int\!e^x\cos x\,dx$

Evidently, you have not been shown the procedure for this type of problem.

We have: . $I \:=\:\int\!e^x\cos x\,dx$

By parts: . $\begin{Bmatrix} u &=& e^x && dv &=& \cos x\,dx \\ du &=& e^x\,dx && v &=& \sin x \end{Bmatrix}$

Then: . $I \;=\;e^x\sinx - \int\!e^x\sin x\,dx$

This integral is as "bad" as the original integral, but watch this!

By parts again: . $\begin{Bmatrix}u &=& e^x && dv &=& \sin x\,dx \\ du &=& e^x\,dx && v &=&\text{-}\cos x \end{Bmatrix}$

We have: . . $I \;=\;e^x\sin x - \left[\text{-}e^x\cos x + \int\!e^x\cos x\,dx\right]$

. . . . . . . . . $I \;=\;e^x\sin x + e^x\cos x - \underbrace{\int\!e^x\cos x\,dx}_{\text{This is }I} \,+\;C$

. . . . . . . . . $I \;=\;e^x(\sin x + \cos x) - I + C$

. . . . . . . . $2I \;=\;e^x(\sin x + \cos x) + C$

Therefore: . $I \;=\;\tfrac{1}{2}e^x(\sin x + \cos x) + C$
July 27th 2012, 06:36 AM
Chuzzle

Re: Integration of Product of e, with variable exponent, and a trigonometric function

Thanks so much Soroban! I would never have considered that if the integral is expanded out to produce extra terms, the second integral of v(du/dx) in the supposedly infinite chain - if you follow me - would be a cosine and therefore the original integral or a factor of it, which can then be substituted; however, I would also never have used v = - cos x for that second integration of v(du/dx), that was true genius. Thanks again for taking the time to help me out, I really appreciate it, you are a true legend made of win. Peace.
July 27th 2012, 06:45 AM
Chuzzle

Re: Integration of Product of e, with variable exponent, and a trigonometric function

Scratch that, only just remembered that the integral of sin x is - cos x . Derp derp derp, that tends to happen at 1 AM. Thanks again!