Evaluate the following Integrals:

Let

As we have the trigonometric identity

to evaluate this integral we use x = acos 2θ

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

When x = –a, acos 2θ = –a ⇒ cos 2θ = –1

When x = a, acos 2θ = a ⇒ cos 2θ = 1

⇒ 2θ = 0 ⇒ θ = 0

So, the new limits are and 0.

Also,

Substituting this in the original integral,

But, we have 2 sin²θ = 1 – cos 2θ