Evaluate the following Integrals:

Let

Note that we can write sin³x = sin²x × sin x

We also have sin²x + cos²x = 1

⇒ sin³x = (1 – cos²x) sin x

So,

Put cos x = t

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

⇒ sin(x)dx = –dt

When x = 0, t = cos 0 = 1

When x = π, t = cos π = -1

So, the new limits are 1 and -1.

Substituting this in the original integral,

Recall