What is the antiderivative of #(1+sinx)/(1-sinx)#?
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#int(1+sin(x))/(1-sin(x))dx=2tan(x)+2/cos(x)-x+C#, #C in RR#
#I=int(1+sin(x))/(1-sin(x))dx#
#=int((1+sin(x))(1+sin(x)))/((1-sin(x))(1+sin(x)))dx#
#=int((1+sin(x))^2)/(1-sin(x)^2)dx#
Because #1-sin(x)^2=cos(x)^2#,
#I=int(sin(x)^2+2sin(x)+1)/(cos(x)^2)dx#
#=int(sin(x)^2)/(cos(x)^2)dx+2intsin(x)/(cos(x)^2)dx+int1/(cos(x)^2)dx#
#=inttan(x)^2dx+2intsin(x)/(cos(x)^2)dx+intsec(x)^2dx#
Now, let #u=cos(x)#
#du=-sin(x)dx#
#I=inttan(x)^2dx+intsec(x)^2dx-2int1/u^2du#
Also, #tan(x)^2=sec(x)^2-1#
#I=2intsec(x)^2dx-int1dx-2int1/u^2du#
#=2tan(x)+2/u-x#
#=2tan(x)+2/cos(x)-x+C#, #C in RR#
\0/ Here's our answer !
