# Evaluate the following : ∫-π4π4x+π42-cos2x⋅dx - Mathematics and Statistics

Sum

Evaluate the following :  int_((-pi)/4)^(pi/4) (x + pi/4)/(2 - cos 2x)*dx

#### Solution

Let I = int_((-pi)/4)^(pi/4) (x + pi/4)/(2 - cos 2x)*dx

= int_((-pi)/4)^(pi/4) [x/(2 - cos2x) + (pi/4)/(2 - cos 2x)]

= int_((-pi)/4)^(pi/4) x/(2 - cos2x)*dx + pi/(4) int_((-pi)/4)^(pi/4) 1/(2 - cos2x)*dx

= "I"_1 + pi/(4)"I"_2                                       ...(1)

Let f(x) = x/(2 - cos2x)

∴ f(– x) = (-x)/(2 - cos[2(-x)]

= (-x)/(2 - cos 2x)

= – f(x)

∴ f is an odd function

∴ int_((-pi)/4)^(pi/4) f(x)*dx = 0

i.e. int_((-pi)/4)^(pi/4) x/(2 - cos 2x)*dx = 0, i.e. I1 = 0     ...(2)

In I2, put tan x = t

∴  x = tan–1t

∴ dx = (1)/(1 + t^2)*dt
and
cos 2x = (1 - t^2)/(1 + t^2)

When x = - pi/(4), t = tan(- pi/4) = – 1

When x = pi/(4), t = tan pi/(4) = 1.

∴ I2 = int_(-1)^(1) (1)/(2 - ((1 - t^2)/(1 + t^2)))*(1)/(1 + t^2)*dt

= int_(-1)^(1) (1)/(2(1 + t^2) - (1 - t^2))*dt

= int_(-1)^(1) (1)/(3t^2 + 1)*dt

= int_(-1)^(1) (1)/((sqrt(3) t)^2 + 1)

= [1/sqrt(3) tan^-1 ((sqrt(3)t)/1)]_(-1)^(1)

= (1)/sqrt(3)[tan^-1 sqrt(3) - tan^-1 (- sqrt(3))]

= (1)/sqrt(3)[tan^-1 sqrt(3) + tan^-1 sqrt(3)]

= (1)/sqrt(3)[pi/3 + pi/3]

= (2pi)/(3sqrt(3)                                              ...(3)
From (1), (2) and (3), we get

I = 0 + pi/(4)[(2pi)/(3sqrt(3))]

= pi^2/(6sqrt(3).

Concept: Fundamental Theorem of Integral Calculus
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Chapter 4: Definite Integration - Exercise 4.2 [Page 172]

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