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Evaluate: Int_0^(Pi/4) "Log" (1 + "Tan" Theta) "D" Theta - Mathematics

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Sum

Evaluate: `int_0^(pi/4) "log" (1 + "tan" theta) "d" theta`

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Solution

Let I = `int_0^(pi/4) "log" (1 + "tan" theta) "d" theta`

Also, I = `int_0^(pi/4) "log"{1 + "tan"(pi/4 - theta)}"d"theta`   
......`[because int_0^a "f"("x") "dx" = int_0^a "f"("a" - "x") "dx"]` 


`= int_0^(pi/4) "log" {1 + ("tan"pi/4 - "tan" theta)/(1 + "tan" pi/4 . "tan" theta)} "d"theta`


`= int_0^(pi/4) "log" {1 + (1 - "tan" theta)/(1 + "tan" theta)}"d" theta`


`= int_0^(pi/4) "log"{(1+"tan" theta + 1 - "tan" theta)/(1+"tan" theta)}"d"theta`


`= int_0^(pi/4) "log" (2/(1+"tan" theta)) "d"theta`


`= int_0^(pi/4) "log"  "2d"  theta - int_0^(pi/4) "log" (1+"tan" theta) "dx"`


`= int_0^(pi/4) "log" 2"d" theta - "I"`


`2"I" = int_0^(pi/4) "log"  2"d"  theta`


`= [theta  "log 2"]_0^(pi/4)`


⇒ 2I = `pi/4` log 2


⇒ I = `pi/8` log 2

Concept: Introduction of Integrals
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