Let I(m,n) = ∫(sin x)^m * (cos x)ⁿ dx

I(m,n) = ∫(sin x)^m * (cos x)^n-1 * cos x dx

Let  u = (sin x)^m * (cos x)^n-1 *

      du = m(sin x)^m-1 *cos x * (cos x)^n-1 + (n - 1)(sin x)^m *(cos x)^n-1 *(-sin x)

=> du = {m(sin x)^m-1 * (cos x)ⁿ - (n - 1)(sin x)^m+1 *(cos x)^n-2 }*dx

Again let dv = cos x dx

Now, integrate on both side, we get,

v = sinx

Now, I(m,n) = (sin x)^m+1 * (cos x)^n-1 - m*I(m,n) - (n - 1)∫(cos x)^{n - 2} * (sin x)^m+2 dx

=> I(m,n) = (sin x)^m+1 * (cos x)^n-1 - m*I(m,n) - (n - 1)∫(cos x)^{n - 2} * (sin x)^m * (sin x)² dx        

=> I(m,n) = {(sin x)^m+1 * (cos x)^n-1 }/(m + n) - (n - 1)/(m + n)∫(cos x)^{n - 2} * (sin x)^m  dx {by putting sin² x = 1 - cos² x and then seperating out terms I(m,n) and rewriting}