# calculus integration problem

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• Sep 9th 2008, 05:58 PM
casesam
calculus integration problem
Indefinite Integral of e^(7x)(sin(4x))dx

cant seem to get the substitution/integration by parts to work for me

thanks in advance!
• Sep 9th 2008, 06:09 PM
Chop Suey
Use integration by parts twice and treat the integral as the unknown. Then you can solve for the integral as if it's an algebra problem.
• Sep 9th 2008, 06:34 PM
casesam
Quote:

Originally Posted by Chop Suey
Use integration by parts twice and treat the integral as the unknown. Then you can solve for the integral as if it's an algebra problem.

im not really sure how to solve it as if the integral were the unknown after 2 integrations by parts, isnt there a way to do this by only using integration by parts and substitution?
• Sep 9th 2008, 08:25 PM
Prove It
Quote:

Originally Posted by casesam
im not really sure how to solve it as if the integral were the unknown after 2 integrations by parts, isnt there a way to do this by only using integration by parts and substitution?

Integration by parts should work, but sometimes one of the most powerful methods of integration is by taking derivatives.

What we are trying to find is

$\displaystyle I=\int (e^{7x}\sin(4x))\,dx$.

If we find the derivative of I, we get

$\displaystyle \frac{d}{dx}(e^{7x}\sin(4x))=4e^{7x}\cos(4x) + 7e^{7x}\sin(4x)$

In reverse, we get that

$\displaystyle \int (4e^{7x}\cos(4x) + 7e^{7x}sin(4x))\,dx=e^{7x}\sin(4x)$
and with a bit of rearranging we find

$\displaystyle 7I= e^{7x}\sin(4x) - 4\int e^{7x}\cos(4x)\,dx$

With me so far?

Now, if we were to divide both sides of the equation by 7, we'd have another expression for I (the function we are integrating). But there's another integral... Let's use the same process to solve for this integral.

Let's call this integral J...

so $\displaystyle 7I=e^{7x}\sin(4x)-4J$ if $\displaystyle J=\int e^{7x}\cos(4x)\,dx$.

Taking the derivative of J, we get

$\displaystyle \frac{d}{dx}(e^{7x}\cos(4x))=-4e^{7x}\sin(4x) + 7e^{7x}cos(4x)$

or in reverse, that

$\displaystyle \int (-4e^{7x}\sin(4x) + 7e^{7x}cos(4x))\,dx = e^{7x}\cos(4x)$.

With a bit of rearranging we find that...

$\displaystyle J=\frac{1}{7}e^{7x}\cos(4x) + \frac{4}{7}\int e^{7x}sin(4x)\,dx$
or
$\displaystyle J=\frac{1}{7}e^{7x}\cos(4x) + \frac{4}{7}I$

Substituting J into our original equation gives...

$\displaystyle 7I= e^{7x}\sin(4x) - 4[\frac{1}{7}e^{7x}\cos(4x) + \frac{4}{7}I]$

which, when expanded and manipulated, reduces to

$\displaystyle \frac{65}{7}I=e^{7x}\sin(4x) - \frac{4}{7}e^{7x}\cos(4x)$

And finally, we can solve for I, which is what we were finding...

$\displaystyle I=\frac{7}{65}e^{7x}\sin(4x)-\frac{4}{65}e^{7x}\cos(4x) + C$ (the integration constant).

Yes, it is a fair bit of writing, but is extremely powerful if all else fails...