Evaluate:

OR

Evaluate:

Using

sin(π – x) = sinx and cos(π – x) = -cosx

Add (i) and (ii)

Substitute cosx = t

When x = 0 ⇒ t = 1 and when x = π ⇒ t = -1

Differentiate t = cosx with respect to x

⇒ -dt = sinxdx

Put in (a)

Using

⇒ I = 2π[tan^-1(1) – tan^-1(-1)]

⇒ I = π²

Hence

OR

Multiply and divide by 2,

⇒ I = 1/2(I₁ – I₂) …(i)

Let us first solve I₁

Substitute x² + 5x + 6 = t

Differentiate with respect to x

⇒ dt = (2x + 5)dx

⇒ I₁ = 2√t

Resubstitute t

Now let us find I₂

Using the result

Put I₁ and I₂ value in (i)