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Evaluate the following integral.

$$ \int e^{2x}\sin{5x}\ dx $$

What I have tried :

$ g(x) = \sin5x , f^{'}(x) = e^{2x} , f(x) = e^{2x} $

$$ \int e^{2x}\sin{5x}\ dx = e^{2x}\sin{5x} -\int e^{2x}\ 5\cos{}5x\ dx $$ $$e^{2x}\sin{5x}-5[e^{2x}\cos{5x}- \int e^{2x}-5\ \sin{5x}\ dx]$$ $$e^{2x}\sin{5x}-5[e^{2x}\cos{5x}- 5\int e^{2x}\sin{5x}\ dx]$$

$$\int e^{2x}\sin{5x}\ dx = e^{2x}\sin5x - 5e^{2x}\cos5x + 25\int e^{2x}\sin5x\ dx$$

$$-24\int e^{2x}\sin5x\ dx = e^{2x}\sin5x-5e^{2x}\cos5x$$

$$ \int e^{2x}\sin5x\ dx = \frac{e^{2x}\sin5x-5e^{2x}\cos5x}{-24} $$

The answer in the book is $ \frac{2}{29}e^{2x}\sin5x-\frac{5}{29}e^{2x}\cos5x+C $

Where is my mistake ?

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  • $\begingroup$ Ever heard of Wolfram Alpha? $\endgroup$
    – J.R.
    Feb 9 '14 at 20:01
  • $\begingroup$ No, What is that ? $\endgroup$ Feb 9 '14 at 20:01
  • $\begingroup$ Click on the link. $\endgroup$
    – J.R.
    Feb 9 '14 at 20:01
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The integral of $e^{2x}$ is, $\int e^{2x}dx = \frac{e^{2x}}{2} + c$. That's your (first) mistake. I didn't look for more.

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  • $\begingroup$ Shouldn't the answer be $ \frac{2}{29}e^{2x}sin5x-\frac{10}{29}e^{2x}cos5x $ ? I did the integral right this time. $\endgroup$ Feb 9 '14 at 20:24
  • $\begingroup$ While I commend your skepticism of the answer in the back of the book (they are frequently wrong), this time it is correct. My best guess for where you made your mistake this time (because making this mistake gives your answer) is that in your second integration by parts, you once again did that $f^\prime(x) = e^{2x}$ and $f(x) =e^{2x}$ where it should be $f(x) = e^{2x}/2$. Be careful ensuring you correctly distribute all your coefficients. EDIT: In one of the two places you substitute in $f(x)$ the second time I believe you made that mistake in order to get that answer. $\endgroup$
    – mlg4080
    Feb 9 '14 at 20:30
  • $\begingroup$ Yeah, I made the same mistake again sorry. Thanks for your help :) $\endgroup$ Feb 9 '14 at 21:08

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