Evaluate the integral. 17 sin2(x) cos3(x) dx step 1 since 17 sin2(x) cos3(x) dx has an odd power of of cos(x), we will convert all but one power to sines.

To evaluate 17 int (sin^2 (x)  cos^3(x))
From Trig identity. Cos^2(x) + sin^2(x) =1. Cos^2(x) = 1 - sin^2 (x) 
Cos^3(x) = cosx * (1 - sin^2 (x)) = cosx - cosxsin^2x
So we have 17 int (sin^2x(cosx - cosxsin^2x))
int (sin^2x(cosx)dx - int (sin^4xcosx)dx. (1)
Let u = sinx then du = cosxdx
Substituting into (1) we have
int (u^2du) - int (u^4du)
u^3/3 - u^5/5
Substitute value for u we have 
(sinx)^3/3 - (sinx)^5/5
Hence we have 17 [ sin^3x/3 - sin^5x/5]

It’s d for sure i just did also that’s why