Question

 Show that : `int _0^(pi/4) "log" (1+"tan""x") "dx" = pi /8 "log"2`

Answer 1

Let I = `int _0^(pi/4) "log" (1+"tan""x") "dx"`

= `int _0^(pi/4) "log" { 1+ "tan"(pi/4-"x")}"dx"`   

 `(because  int _0^"a" "f"("x") "dx" = int  "f" ("a" -"x") "dx") `

= `int _0^(pi/4) "log" {1+(("tan" pi /4  -"tan""x"))/(1+ "tan" pi/4"tan""x" }}"dx"`

`=int _0^(pi/4) {1+(1-"tan""x")/(1+"tan""x")} "dx"`

`=int _0^(pi/4) "log"{(1+"tan""x" + 1-"tan""x")/(1 + "tan""x")} "dx"`

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

`=int_0^(pi/4) {"log"2 -"log"(1+"tan""x")}"dx" `

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

`"I" = "log"2["x"]_0^(pi/4) - "I"`

`2"I" = "log"2[pi/4-0]`

`"I" = pi /8 . "log"2`

`therefore int_0^(pi/4) "log"(1+ "tan""x")"dx" = pi/8"log"2`