Evaluate:

Let I = …(1)

As the expression is bit complicated and none of the property of definite integral is going to help in it.

So we need to integrate it simply and put the limits.

∵ I =

⇒ I =

∴ I = I₁ + I₂

Where I₁ = and I₂ =

Evaluation of I₁:

∵ I₁ =

∵ 1 + cos x = 2cos² (x/2)

∴ I₁ =

⇒ I₁ =

Applying Integration by parts –

I₁ = {∵}

⇒ I₁ = {∵}

⇒ I₁ =

⇒ I₁ =

Evaluation of I₂:

I₂ =

Let u = 1 + cos x

⇒ du = -sin x dx

When x = 0 ⇒ u = 1 + cos 0 = 2

And when x = π/2 ⇒ u = 1 + cos π/2 = 1

∴ I₂ =

⇒ I₂ =

⇒ I₂ = -[log 1 – log 2] = log 2

∴ I = I₁ + I₂ = + log 2

∵ 2 log √2 = log (√2)² = log 2

∴ I = π/2 – log 2 + log 2

⇒ I = π/2

∴