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Question
Evaluate the following:
`int_0^(pi/2) log(tanx)dx`
Solution
Let I = `int_0^(pi/2) log(tanx)dx`
We use the property, `int_0^a f(x)dx = int_0^a f(a - x)dx`
Here, `a = pi/(2)`
Hence, changing x by `pi/(2) - x`, we get
I = `int_0^(pi/2) log[tan(pi/2 - x)]dx`
= `int_0^(pi/2) log(cotx)dx`
= `int_0^(pi/2) log(1/tanx)dx`
= `int_0^(pi/2) log(tanx)^-1dx`
= `int_0^(pi/2) - log(tanx)dx`
= `- int_0^(pi/2) log(tanx)dx`
= – I
∴ 2I = 0
∴ I = 0
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