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Question
Evaluate: `int_0^(π/4) log(1 + tanx)dx`.
Solution
Let I = `int_0^(π/4) log_e (1 + tan x)dx` ...(i)
`\implies` I = `int_0^(π/4) log_e (1 + tan(π/4 - x))dx`,
Using `int_0^a f(x)dx = int_0^a f(a - x)dx`
`\implies` I = `int_0^(π/4) log_e (1 + (1 - tanx)/(1 + tanx))dx`
= `int_0^(π/4) log_e (2/(1 + tanx))dx`
= `int_0^(π/4) log_e 2dx - I` ...(Using ...(i))
`\implies` 2I = `π/4 log_e 2`
`\implies` I = `π/8 log_e 2`.
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