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

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Answer 1 · 2018-08-12T16:49:18

$int(1+sin(x))/(1-sin(x))dx=2tan(x)+2/cos(x)-x+C$ , $C in RR$

$I=int(1+sin(x))/(1-sin(x))dx$

$=int((1+sin(x))(1+sin(x)))/((1-sin(x))(1+sin(x)))dx$

$=int((1+sin(x))^2)/(1-sin(x)^2)dx$

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 !

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