# Calculus

I am trying to find the integral of e^(6x)sin(7x). Apparently, the answer is (6e^(6x)sin(7x) - 7e^(6x)cos(7x))/85) + C and achieving the answer is mostly understandable since it involves integrating the function in parts... that is until I get to this part:

(e^(6x)sin(7x)/6) - ((7e^(6x)cos(7x)/36) + (49/36 int\_e^(6x)sin(7x))).

It has left me confused on how they even got that answer; integral-calculator.com says to solve for it since it appears again while symbolab.com simply says to isolate the integral. If I remember correctly, 36 and 6 are not even divisible to 85. How do I even achieve the given answer if I don't know what to do from there?

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1. You are good so far, almost there!
one little slip up.

so far you have:
∫e^(6x)sin(7x) dx = (e^(6x)sin(7x)/6) - [(7e^(6x)cos(7x)/36) + 49/36 ∫e^(6x)sin(7x) dx ]
.... for that last part of 49/36 ∫e^(6x)sin(7x) dx , I had ( - 49/36 ∫e^(6x)sin(7x) dx ) . Check your signs carefully

then ...
∫e^(6x)sin(7x) dx = (e^(6x)sin(7x)/6) - (7e^(6x)cos(7x)/36) - 49/36 ∫e^(6x)sin(7x) dx ] , notice the integrals at the left and at the right are the same!
move that term to the left and add them
(1+49/36)∫e^(6x)sin(7x) dx = (e^(6x)sin(7x)/6) - (7e^(6x)cos(7x)/36)
(85/36) ∫e^(6x)sin(7x) dx = (e^(6x)sin(7x)/6) - (7e^(6x)cos(7x)/36)
times 36
85 ∫e^(6x)sin(7x) dx = 6(e^(6x)sin(7x) - 7e^(6x)cos(7x)
finally divide by 85 and you got it.

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