Maharashtra State BoardHSC Arts 12th Board Exam
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Evaluate the following : ∫-π4π4x+π42-cos2x⋅dx - Mathematics and Statistics

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Sum

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

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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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