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