Evaluate the following pls helpIntegration sin3xcos5x dx

We know that sin(A+B) = sinAcosB + cosAsinB -----------------1
We know that sin(A - B) = sinAcosB - cosAsinB -----------------2

Now adding 1 and 2

2sinAcosB = sin(A+B) + sin(A - B), so that 

sinAcosB = (1/2)[sin(A+B) + sin(A-B)] 

let A = 3x and B = 5x 

Rewriting the question 

(1/2) ∫ [sin8x + sin(-2x)] dx = 

(1/2) ∫ [sin8x - sin2x] dx = 

=> -1/16 cos8x + 1/4 cos2x + C