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tanx-e-ipi-dx-




Question Number 100097 by Dwaipayan Shikari last updated on 24/Jun/20
∫(tanx)^e^(iπ)  dx
(tanx)eiπdx
Commented by Dwaipayan Shikari last updated on 24/Jun/20
∫(tanx)^(−1) dx  ∫cotxdx=log(sinx)+Constant{As  e^(iπ) =−1}
(tanx)1dxcotxdx=log(sinx)+Constant{Aseiπ=1}

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