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What is the antiderivative of #(1+sinx)/(1-sinx)#?

Answer 1

Isabella Ackerman

#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#

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Answer 2

Axel Alvarez

The antiderivative of (\frac{1+\sin(x)}{1-\sin(x)}) is (-\ln|1-\sin(x)| + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.