1. ## Stupid Integral Help

I need to integrate $\int5e^{x}sin\left(2x \right)dx$, and I just cannot figure out what the trick is to it. I don't think it's any combination of integration by parts since $e^{x}$ will always stay the same and the other function will oscillate back and forth between sin and cosine, and all that will happen is that the factors of 5 will increase.

I also tried the double angle identity $sin(2x)=2sin(x)cos(x)$, and that resulted in the same kind of loop. I'm sure it's some little trick that I'm missing here, so any hints would be greatly appreciated.

2. Originally Posted by davesface
I need to integrate $\int5e^{x}sin\left(2x \right)dx$
Go to this website Wolfram|Alpha
In the input window, type in this exact expression: integrate 5e^x sin[2x]dx (you can copy & paste)
Click the equals bar at the right-hand end of the input window.
Be sure you click ‘show steps’ to see the solution
.

3. Okay, that did give me the answer, but the intermediate step didn't really help out at all. For reference, here is what it showed me:

Is it really only possible to solve this with that formula memorized? Maybe calculus has made me paranoid, but it really does seem like there should be a trick involved somewhere in the process. If anyone can confirm/disprove this, please let me know. If I'm just bad at memorization, then at least thanks for the answer.

4. Originally Posted by davesface
I need to integrate $\int5e^{x}sin\left(2x \right)dx$, and I just cannot figure out what the trick is to it. I don't think it's any combination of integration by parts since $e^{x}$ will always stay the same and the other function will oscillate back and forth between sin and cosine, and all that will happen is that the factors of 5 will increase.

I also tried the double angle identity $sin(2x)=2sin(x)cos(x)$, and that resulted in the same kind of loop. I'm sure it's some little trick that I'm missing here, so any hints would be greatly appreciated.
A number of approaches are possible, including repeated integration by parts - you end up with an equation where the integral I is the unknown and can therefore be made the subject. Read example 5 here: http://www.math.hawaii.edu/~chasse/m242/parts.pdf

5. Originally Posted by davesface
Okay, that did give me the answer, but the intermediate step didn't really help out at all. For reference, here is what it showed me:
Is it really only possible to solve this with that formula memorized?
You have got to realize that The Calculus is not fir the lazy.

6. Originally Posted by Plato
You have got to realize that The Calculus is not fir the lazy.
Plato, I'm not lazy. Laziness would be going to an online integrator, plugging in the question, and then copying down the answer without wondering if there isn't a more human-friendly way to find it (computers can store every possible formula for assisting with integration whereas I cannot, nor should I be expected to). I don't appreciate the condescension, especially considering that my intuition proved to be correct (as in the example supplied by mr fantastic).

To mr fantastic- After reading the example, it was trivial to apply it to my specific problem and get the right answer, as verified by my textbook. Thank you very much for the tip, which I'm sure will come in handy again in the future.

7. Then all's well that ends well. Thread closed.