# 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

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