Question

What is `f(x) = int xsin2x+cos3x dx` if `f(pi/12)=4`?

Answer 1

`f(x) = -x/2cos(2x)+1/4sin(2x)+1/3 sin(3x) + 1/48 (186 - 8 sqrt(2) + sqrt(3) pi)`

Explanation:

Given: `f(x) = int xsin(2x)+cos(3x) dx; f(pi/12) = 4`

`f(x) = int xsin2xdx+int cos3x dx`

Setup the integration by parts variable for the first integral:

Let `u = x and dv = sin(2x)dx`
Then `du = dx` and `v = -1/2cos(2x)dx`

`f(x) = -x/2cos(2x)+1/2int cos(2x)dx+int cos3x dx`

The remaining integrals become sine functions multiplied by a constant:

`f(x) = -x/2cos(2x)+1/4sin(2x)+1/3 sin(3x) + C`

To find the value of C, evaluate at the point `(pi/12,4)`

`4 = -(pi/12)/2cos(2(pi/12))+1/4sin(2(pi/12))+1/3 sin(3(pi/12)) + C`

I let WolframAlpha sort this one out:

`C = 1/48 (186 - 8 sqrt(2) + sqrt(3) pi)`

Substitute the value for C into `f(x)`:

`f(x) = -x/2cos(2x)+1/4sin(2x)+1/3 sin(3x) + 1/48 (186 - 8 sqrt(2) + sqrt(3) pi)`