integration problem: [int]e^(7x)sin(7x)dx

[dmsjr12]

I ran into this problem during my homework.

[int]e^(7x)sin(7x)dx

I used integration by parts.

u=sin(7x)
du=7cos(7x)
v=1/7 e^(7x)
dv=e^(7x)

i believe i have to do integration by parts twice, but when i tried it I couldn't get anything that made sense. Can someone please help me?

 

[galactus]

Let \(\displaystyle \L\\u=e^{7x}, \;\ dv=sin(7x), \;\ v=\frac{-cos(7x)}{7}, \;\ du=7e^{7x}\)

You get:

\(\displaystyle \L\\\frac{-e^{7x}cos(7x)}{7}+\int[e^{7x}cos(7x)]dx\)

Now, use integration by parts again using the subs:

\(\displaystyle \L\\u=e^{7x}, \;\ dv=cos(7x)dx, \;\ du=7e^{7x}, \;\ v=\frac{sin(7x)}{7}\)

You will end up with the same integral you started with. Add it to both sides and divide through by 2.

Let me know what you get.