Let

In the denominator, we have sin 2x = 2 sin x cos x

Note that we can write 2 sin x cos x = 1 – (1 – 2 sin x cos x)

We also have sin²x + cos²x = 1

⇒ 1 – 2 sin x cos x = sin²x + cos²x – 2 sin x cos x

⇒ sin 2x = 1 – (sin x – cos x)²

So, using this, we can write our integral as

Now, put sin x – cos x = t

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

When x = 0, t = sin 0 – cos 0 = 0 – 1 = -1

When,

So, the new limits are -1 and 0.

Substituting this in the original integral,

Recall,