Indefinite Integral of e^(7x)(sin(4x))dx

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

thanks in advance!

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- Sep 9th 2008, 05:58 PMcasesamcalculus 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 PMChop Suey
Use integration by parts

and treat the integral as the unknown. Then you can solve for the integral as if it's an algebra problem.*twice* - Sep 9th 2008, 06:34 PMcasesam
- Sep 9th 2008, 08:25 PMProve It
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...