Finding $$\int e^{2x} \sin 4x \, dx$$
I think I should be doing integration by parts…
If I let $u=e^{2x} \Rightarrow du = 2e^{2x}$,
$dv = \sin{4x} \Rightarrow v = -\frac{1}{4} \cos{4x}$
$\int{ e^{2x} \sin{4x}} dx = e^{2x}(-\frac{1}{4}\cos 4x) – \color{red}{\int (-\frac{1}{4} \cos 4x) 2e^{2x} \, dx}$
Highlighted in red, I seem to integrating an exponential times trig expression again … doesn't seem like te right way to go
So I let $u=\sin{4x} \Rightarrow du = 4\cos 4x \, dx$
$dv = e^{2x} dx \Rightarrow v = \frac{1}{2} e^{2x}$
$\sin{4x}(\frac{1}{2} e^{2x}) – \int (\frac{1}{2} e^{2x})(4\cos4x) \, dx$
Again, its an exponential times a trig function?
Am I using the wrong substitution?
Best Answer
I wonder how many times this same question has been asked here, and how many times it is asked every semester in every second-semester calculus course?
If you integrate by parts a second time, you get the same integral you started with, and the naive reaction is that that means you're getting nowhere. But in truth it means you're almost done. You've got $$ \int \text{something} = \text{something} - \text{something}\cdot \int [\text{same integral}]. $$ So you add the same thing to both sides of the equation and get $$ \int \text{something} + \text{something}\cdot\int [\text{same integral}] = \text{something} + C $$ Then you write $$ (1+\text{something}) \cdot\int\cdots\cdots = \text{something} + C $$ $$ \int\cdots\cdots = \frac{\text{something}}{1+\text{something}} + \text{constant}. $$