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What's the integral of ##int [ ( secx ) / ( 1+ tanx ) ] ^2 dx##?
##-1/(1 + tanx) + c##
In order to do this integral, let's square the top and the bottom, we get:
##int sec^2(x)/(1 + tanx) ^2 dx##
Since the derivative of ##tanx## is ##sec^2(x)##, we can do a u-substitution
##int sec^2(x)/(1 + tanx)^2 dx##
##u = 1 + tanx##
##du = sec^2(x) dx##
The new integral would be:
##int 1/u ^2 du##
This turns out to be ##-1/u + c##, and then we put back the ##1 + tanx## and get ##-1/(1 + tanx) + c##.