Indefinite Integral of e^(7x)(sin(4x))dx
cant seem to get the substitution/integration by parts to work for me
thanks in advance!
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...