Let

We have sin²x + cos²x = 1

Put sin²x = t

⇒ 2 sin x cos x dx = dt (Differentiating both sides)

⇒ sin x cos x dx = dt

When x = 0, t = sin²0 = 0

So, the new limits are 0 and 1.

Substituting this in the original integral,

Recall