Question

any help?
Evaluate the integral ∫e^2t sin(-5t) dt ?

Answer 1

\[\int\limits e^(2t) \sin(-5t) dt\]

Answer 2

This will be a "repeating" integral. Integrate by parts.

Answer 3

i tried that it didnt work?

Answer 4

Use Euler's equation to simplify if you've covered that.

Answer 5

By a "repeating" integral, what Turing means is that integrate by parts (the integral that you see when you integrate the first time) a second time. When you integrate by parts a second time, you'll notice that the integrals eventually cancel and you get something nice.

Answer 6

\[\Im(\int\limits e^{(2-5i)t}dt)=\Im(\frac{1}{2-5i}e^{(2-5i)t})\]

Answer 7

i did it this way: \[e^(2t) \int\limits \sin(-5t) dt \]

Answer 8

t is not a constant so you cannot take it out of the integral
do you know how to integrate by parts?

Answer 9

but its 2xt?

Answer 10

Whether it's 2 times t or just t, t is a variable, not a constant. You cannot remove \[e ^{2t}\] from the integral as a constant since you are integrating with respect to t.

Answer 11

i still cant get it?

Answer 12

Here is what you are looking for, a step-by-step calculation, using integration by parts two times, successively: http://mathforum.org/library/drmath/view/53749.html

Answer 13

still confused?

Answer 14

\[\int e^{2t}\sin(-5t)\;dt\]
Since sine is an odd function, you have sin(-5t) = -sin(5t)
\[-\int e^{2t}\sin(5t)\;dt\]
Integrating by parts, let
\[\begin{matrix}u=e^{2t}& & dv=\sin(5t)\;dt\\
du=2e^{2t}dt& & v=-\frac{1}{5}\cos(5t)\end{matrix}\\
\begin{align*}-\int e^{2t}\sin(5t)\;dt&=-\frac{1}{5}e^{2t}\cos(5t)-\int\left(-\frac{1}{5}\cos(5t)\right)\left(2e^{2t}dt\right)\;dt\\
&=-\frac{1}{5}e^{2t}\cos(5t)+\frac{2}{5}\int e^{2t}cos(5t)\;dt\end{align*}\]
For the next integral, repeat integration by parts:
\[\begin{matrix}f=e^{2t}& & dg=\cos(5t)\;dt\\
df=2e^{2t}dt& & g=\frac{1}{5}\sin(5t)\end{matrix}\\
\begin{align*}-\int e^{2t}\sin(5t)\;dt&=-\frac{1}{5}e^{2t}\cos(5t)+
\\&\;\;\;\;\;\;\;\;\;\; \frac{2}{5}\left[\frac{1}{5}e^{2t}\sin(5t)-\int\left(\frac{1}{5}\sin(5t)\right)\left(2e^{2t}\right)\;dt\right]\\
\color{red}{-\int e^{2t}\sin(5t)\;dt}&=-\frac{1}{5}e^{2t}\cos(5t)+
\\&\;\;\;\;\;\;\;\;\;\; \frac{2}{25}e^{2t}\sin(5t)\color{red}{-\frac{4}{25}\int e^{2t}\sin(5t)\;dt}\end{align*}\]
The red text is the repetition @TuringTest was referring to.

Answer 15

so if i work that out it should be the answer? ?

Answer 16

Yes. Moving the integral on the RHS to the LHS, then simplifying until you have
\[\int e^{2t}\sin(5t)\;dt=\cdots\]
Of course, you're supposed to find the antiderivative of e^(2t) sin(-5t), so make sure you rewrite in terms of that.