Question

How do you integrate `sin^5 (x) * cos^3 (x)`?

Answer 1

Use one of the substitutions: `u=sinx` `" "` OR `" "` `u=cosx`

Explanation:

`int sin^5x cos^3x dx = int sin^5xcos^2xcosx dx`

`= int sin^5x(1-sin^2x)cosx dx`

`= int sin^5x cosx dx - int sin^7xcosx dx`

Substitute `u = sinx` to get

`= 1/6 sin^6x - 1/8 sin^8x +C`

OR

`int sin^5x cos^3x dx = int sin^4xcos^3xsinx dx`

`= int (sin^2x)^2 cos^3x sinx dx`

`= int (1-cos^2x)^2 cos^3x sinx dx`

`= int (1-2cos^2x+cos^4x) cos^3x sinx dx`

`= int cos^3x sinx dx - int 2cos^5x sinx dx + int cos^7x sinx dx`

Substitute `u = cosx`, so `du = -sinx` to get

`= -1/4cos^4x+1/3cos^6x-1/8cos^8x +C`