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Evaluate: π∫0π4log(1+tanx)dx.

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Question

Evaluate: `int_0^(π/4) log(1 + tanx)dx`.

Sum
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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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2023-2024 (March) Board Sample Paper

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