Let

As we have the trigonometric identity

to evaluate this integral we use x² = a²cos 2θ

⇒ 2xdx = –2a²sin(2θ)dθ (Differentiating both sides)

⇒ xdx = –a²sin(2θ)dθ

When x = a, a²cos 2θ = a²⇒ 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