Let

Put cos^-1x = t

⇒ x = cos t

⇒ dx = – sin t dt (Differentiating both sides)

When x = 1, t = cos^-1(1) = 0

So, the new limits are and 0.

Substituting this in the original integral,

We will use integration by parts.

Recall

Here, take f(t) = t² and g(t) = sin t

Now,

⇒ f’(t) = 2t

Substituting these values, we evaluate the integral.

Let

We use integration by parts again.

Here, take f(t) = t and g(t) = cos t

Now,

⇒ f’(t) = 1

Using these values in equation for I₁

Substituting I₁ in I, we get

⇒ I = –2 – (–π) = π – 2