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Question
Evaluate `int_0^(π//4) log (1 + tanx)dx`.
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Solution
Let I = `int_0^(π//4) log (1 + tanx)dx` ...(i)
By using property
`int_0^a f(x) = int_0^a f(a - x)`
I = `int_0^(π//4) log [1 + tan(π/4 - x)]dx`
= `int_0^(π//4) log [1 + (tan π/4 - tan x)/(1 + tan π/4 tan x)]dx`
= `int_0^(π//4) log [1 + (1 - tanx)/(1 + tanx)]dx`
= `int_0^(π//4) log [2/(1 + tanx)]dx`
= `int_0^(π//4) log 2 - int_0^(π//4) log (1 + tanx)dx` ...(ii)
On adding equations (i) and (ii),
2I = `int_0^(π//4) log (1 + tanx)dx + int_0^(π//4) log 2 dx - int_0^(π//4) log (1 + tanx)dx`
`\implies` 2I = `int_0^(π//4) log 2 dx`
`\implies` 2I = `log 2 int_0^(π//4) 1.dx`
`\implies` 2I = `log 2 [x]_0^(π//4)`
`\implies` I = `log2/2 xx π/4 = π/8 log 2`
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