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

1 Answer
Oct 28, 2017

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)