Question

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

Answer 1

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