Evaluate the following Integrals:

Let

As we have the trigonometric identity, to evaluate this integral we use x = cos 2θ

⇒ dx = –2sin(2θ)dθ (Differentiating both sides)

When x = 0, cos 2θ = 0 ⇒ 2θ = ⇒ θ =

When x = 1, cos 2θ = 1 ⇒ 2θ = 0 ⇒ θ = 0

So, the new limits are and 0.

Substituting this in the original integral,

We have and sin 2θ = 2 sin θ cos θ

But,