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Question
Show that : `int _0^(pi/4) "log" (1+"tan""x")"dx" = pi /8 "log"2`
Solution
Let I = `int _0^(pi/4) "log"(1+"tan""x")"dx"`
= `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) "log"{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"]int _0^(pi/4) - "I"`
2I = `"log" 2 [pi/4-0]`
`"I" = pi/8 ."log"2`
` therefore int _0^(pi/4) "log"(1 +"tan""x")"dx" = pi/8"log"2`
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