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Evaluate: int0^(pi//2)logtanx\ dx...

Evaluate: `int_0^(pi//2)logtanx\ dx`

Text Solution

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Let I=`int_0^(pi//2)logtanxdx`
I_1=`int_0^(pi//2)logtan(pi/2-x) dx`
I_1=`int_0^(pi//2)logcotx dx`
I_1=`-int_0^(pi//2)logtanx dx`
`I=−I_1`
⟹I=0
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