Let

We have sin²x + cos²x = 1 and sec²x – tan²x = 1

We can write sin³x = sin²x × sin x = (1 – cos²x) sin x

Put cos x = t

⇒ –sin(x)dx = dt (Differentiating both sides)

⇒ sin(x)dx = –dt

When x = 0, t = cos 0 = 1

So, the new limits are 1 and 0.

Substituting this in the original integral,

Recall