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?
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.
To find the integral of e^(6x)sin(7x), you can use integration by parts. Let's go step-by-step to understand how to get to the given answer:
1. Apply integration by parts, using the formula:
∫u * dv = u * v - ∫v * du
Let u = e^(6x) and dv = sin(7x)dx, which implies du = 6e^(6x)dx and v = (-1/7)cos(7x).
2. Use the integration by parts formula to find the first part of the integral:
∫e^(6x)sin(7x)dx = -e^(6x)cos(7x)/7 - ∫(-1/7)cos(7x) * 6e^(6x)dx
Simplifying the above equation yields:
∫e^(6x)sin(7x)dx = -e^(6x)cos(7x)/7 + (6/7) ∫e^(6x)cos(7x)dx
3. We now have a similar integral on the right side of the equation: ∫e^(6x)cos(7x)dx.
To solve this second integral, repeat steps 1 and 2 by setting u = e^(6x) and dv = cos(7x)dx. You will find that:
∫e^(6x)cos(7x)dx = (e^(6x)sin(7x))/7 - (6/7)∫e^(6x)sin(7x)dx
4. Substitute the second integral back into the first equation:
∫e^(6x)sin(7x)dx = -e^(6x)cos(7x)/7 + (6/7) * [(e^(6x)sin(7x))/7 - (6/7)∫e^(6x)sin(7x)dx]
Now, let's simplify the equation step-by-step:
Multiply (6/7) into each term within the square brackets:
= -e^(6x)cos(7x)/7 + (6/49) * e^(6x)sin(7x) - (36/49) ∫e^(6x)sin(7x)dx
5. Move the (-36/49) times the integral term to the left side of the equation:
(1 + (36/49)) ∫e^(6x)sin(7x)dx = -e^(6x)cos(7x)/7 + (6/49) * e^(6x)sin(7x)
= (85/49) ∫e^(6x)sin(7x)dx = -e^(6x)cos(7x)/7 + (6/49) * e^(6x)sin(7x)
6. Divide both sides of the equation by (85/49):
∫e^(6x)sin(7x)dx = (-e^(6x)cos(7x))/(7 * (85/49)) + ((6/49) * e^(6x)sin(7x))/(85/49)
Simplifying the above equation gives:
∫e^(6x)sin(7x)dx = (-e^(6x)cos(7x))/(7 * (85/49)) + ((6/49) * e^(6x)sin(7x))/(85/49)
= (6e^(6x)sin(7x) - 7e^(6x)cos(7x))/85 + C
So, the integral of e^(6x)sin(7x) is indeed (6e^(6x)sin(7x) - 7e^(6x)cos(7x))/85 + C, as provided.
To find the integral of a function, we often use integration techniques like integration by parts. In this case, you correctly started by applying integration by parts. However, it seems you got confused when you reached the step:
(e^(6x)sin(7x)/6) - ((7e^(6x)cos(7x)/36) + (49/36 int_e^(6x)sin(7x)))
To simplify this expression further, we can rewrite it as:
(e^(6x)sin(7x)/6) - ((7e^(6x)cos(7x) + 49/36 int_e^(6x)sin(7x)) / 36)
Now, let's combine the two terms involving the integral:
(e^(6x)sin(7x)/6) - (7e^(6x)cos(7x) + 49/36 int_e^(6x)sin(7x)) / 36
To isolate the integral term, let's multiply the entire equation by 36:
36(e^(6x)sin(7x)/6) - (7e^(6x)cos(7x) + 49/36 int_e^(6x)sin(7x)) = 0
Simplifying further:
6*6e^(6x)sin(7x) - 7*36e^(6x)cos(7x) - 49 int_e^(6x)sin(7x) = 0
Rearranging the equation:
36e^(6x)sin(7x) - 252e^(6x)cos(7x) - 49 int_e^(6x)sin(7x) = 0
Now, let's isolate the integral term:
49 int_e^(6x)sin(7x) = 36e^(6x)sin(7x) - 252e^(6x)cos(7x)
Finally, divide both sides by 49 to solve for the integral:
int_e^(6x)sin(7x) = (36e^(6x)sin(7x) - 252e^(6x)cos(7x))/49
At this point, you can simplify the expression further if you'd like:
int_e^(6x)sin(7x) = (6e^(6x)sin(7x) - 42e^(6x)cos(7x))/7
However, please note that the given answer you mentioned in your original question is different from the final result we derived. If you follow the steps I provided, you would reach the expression above rather than the one you mentioned.