Question

∫e^(2t)sin(3t)dx, integral, can someone help me and explain with details!

Answer 1

So, what happens with this kind of an integral is we eventually end up in a never-ending loop of integration by parts. When we have a function that does this, we actually have something we can do after two integrations that will help us solve it. So let's do integration by parts twice and Ill show ya what I mean. So this is the integration by parts formula:

\[u \int\limits_{}^{}v-[\int\limits_{}^{}u'*\int\limits_{}^{}v]\]So we choose u to be a portion of our function that is easy or convenient to integrate and our v to be a portion of our function that is easy to integrate. In problems liek this, we want to choose u to be the trig portion and v to be the e portion. So u = sin3t, u' = 3cos3t, v = e^(2t), integral v = (1/2)e^(2t). So now I just plug in those pieces into the formula to get:
\[\frac{ e^{2t} }{ 2 }\sin(3t) - \frac{ 3 }{ 2 } \int\limits_{}^{}e ^{2t} \cos(3t) \]The 3/3 is just factored out of the integral. Now I have done by parts once, but Ill need to do it once more. This time u will be cos(3t), u' = -3sin(3t) and v is still e^(2t) with integral v being (1/2)e^(2t). So plugging those in and doing by parts a second time I get:
\[\frac{ e^{2t} }{ 2 }\sin(3t)-\frac{ 3 }{ 2 }[\frac{ e ^{2t} }{ 2 }\cos(3t) +\frac{ 3 }{ 2 } \int\limits_{}^{}e^{2t}\sin(3t)]\]
So now I have done the required integration by parts liek I said before. So now Im going to multiply that -3/2 through to get:
\[\int\limits_{}^{}e^{2t}\sin(3t)dt = \frac{ e^{2t} }{ 2 }\sin(3t)-\frac{ 3e^{2t} }{ 4 }-\frac{ 9 }{ 4 }\int\limits_{}^{}e ^{2t}\sin(3t)\]Now here is where the trick comes in. We start out on the left iwth integral e^(2t)sin(3t) and on the other side of the equation we have -9/4integal e^(2t)sin(3t). These are actually like terms. I can add (9/4)integral e^2t)sin(3t) to both sides like I would in a normal algebra equation. So on the left I have 1 integral e^2tsin(3t) and Ill be adding to both sides (9/4)integral e^(2t)sin(3t).
So if I do that addition I end up with this:

\[\frac{ 13 }{ 4 }\int\limits_{}^{}e^{2t}\sin(3t)dt =\frac{ e^{2t} }{ 2 }\sin(3t)-\frac{ 3e^{2t} }{ 4 }\cos(3t)\]So since the original problem wanted me to solve for integral e^2tsin(3t), Ill multiply both sides by 4/13 to get the integral by itself and essentially solve for it. After I do that, my final answer is:

\[\frac{ 4 }{ 13 }(\frac{ e^{2t} }{ 2 }\sin(3t)-\frac{ 3e^{2t} }{ 4 }\cos(3t)) + C\]