Answer :

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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