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Prove that int0^(pi//8) log |1 + tan 2x|...

Prove that `int_0^(pi//8) log |1 + tan 2x|\ dx = pi/16 log_e 2.`

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`"Let I "= int_(0)^(pi//8) log (1+tan 2x) dx " "....(1)`
` rArr I=int_(0)^(pi//8) log { 1+tan 2((pi)/(8)-x)} dx`
`=int_(0)^(pi//8) log {1+tan ((pi)/(4)-2x)}dx`
`=int_(0)^(pi//8) log { 1+((1-tan 2x)/(1+tan 2x)) }dx`
` =int_(0)^(pi//8) log ((2)/(1+tan 2x))dx`
`=int_(0)^(pi//8) log((2)/(1+tan 2x)) dx`
` =int_(0)^(pi//8) log 2 dx - int_(0)^(pi//8) log (1 +tan 2x) dx`
`rArr I= log 2[x]_(0)^(pi//8) -I`
`rArr 2I= (pi)/(8) log 2`
` rArr I=(pi)/(16) log 2`
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