Question

Find the integrals of the function:

sin³ (2x + 1)

Answer 1

Let `I = int sin^3 (2x + 1)` dx

`= 1/4 [3 sin (2x + 1) - sin 3 (2x + 1)]` dx

        .....`[because sin^3 theta = 1/4 (3 sin theta - sin 3 theta)]`

`= 3/4 (- (cos (2x + 1))/2 - 1/4 ((- cos 3 (2x + 1))/6) + C`

`= -3/8  cos (2x + 1) + 1/24  cos 3 (2x + 1) + C`

`= - 3/8  cos (2x + 1) + 1/24 [4 cos^3 (2x + 1) - 3 cos (2x + 1) + C`      ......`[because  cos 3 theta = 4 cos^3 theta - 3 cos theta]`

`= - 3/8  cos (2x + 1) + 1/6  cos^3 (2x + 1) - 1/8  cos(2x + 1) + C`

`= - 1/2  cos (2x + 1) + 1/6  cos^3 (2x + 1) + C`