Question

Integration problem...

Answer 1

\[\int\limits\limits_{}^{}e^{-3x}\sin(5x)dx\]
I started integrating by parts:\[u = \sin(5x)\]\[dv - e^{-3x}dx\]\[du = 5\cos(5x)dx\]\[v=-\frac{1}{3}e^{-3x}\]
So...
\[\int\limits\limits_{}^{}e^{-3x}\sin(5x)dx = -\frac{1}{3}e^{-3x}\sin(5x)+\frac{5}{3}\int\limits\limits_{}^{}e^{-3x}\cos(5x)dx\]
Then a 2nd integration by parts:
\[u = \cos(5x)\]\[dv=e^{-3x}dx\]\[du=-5\sin(5x)dx\]\[v=-\frac{1}{3}e^{-3x}\]
So...\[ = -\frac{1}{3}e^{-3x}\sin(5x)+\frac{5}{3} \left[ -\frac{1}{3}e^{-3x}\cos(5x)+\frac{5}{3}\int\limits\limits e^{-3x}\sin(5x)dx \right]\]\[ = -\frac{1}{3}e^{-3x}\sin(5x) -\frac{5}{9}e^{-3x}\cos(5x)+\frac{25}{9}\int\limits\limits e^{-3x}\sin(5x)dx\]By Subracting the integral from both sides, you get\[-\frac{16}{9}\int\limits\limits e^{-3x}\sin(5x)dx = -\frac{1}{3}e^{-3x}\sin(5x) -\frac{5}{9}e^{-3x}\cos(5x)\]Then\[ \int\limits\limits\limits e^{-3x}\sin(5x)dx = \frac{3}{16}e^{-3x}\sin(5x) +\frac{5}{16}e^{-3x}\cos(5x)\]

However, wolfram alpha says it should be:\[ \int\limits\limits\limits\limits e^{-3x}\sin(5x)dx = -\frac{3}{34}e^{-3x}\sin(5x) -\frac{5}{34}e^{-3x}\cos(5x) \]
Where did I go wrong?

Answer 2

\[= -\frac{1}{3}e^{-3x}\sin(5x)+\frac{5}{3} \left[ -\frac{1}{3}e^{-3x}\cos(5x)-\frac{5}{3}\int\limits\limits\limits e^{-3x}\sin(5x)dx \right] \]
*
I think...
I should write this on paper first

Answer 3

In your second integration by parts, you should have
\[ ---5/3 \int e^{-3x} \sin(5x) \ dx \]
In other words, you dropped a minus sign

Answer 4

:) yes what jamesj said!

Answer 5

(one minus from the e^-3x, one from the derivative of cos(5x), and one from the integration by parts. Classic mistake; we all make it.)

Answer 6

Ah, you're right! Thanks :)