`inttanx/(1-sinx)`?

`I=int(tanx)/(1-sinx)dx`

Divide each, numerator and denominator by `cosx`

`=int(tanx/cosx)/(1/cosx-sinx/cosx)dx`

`=int(secxtanx)/(secx-tanx)dx`

Multiply and divide the expression by `(secx+tanx)`

`=int((secxtanx)(sec+tanx))/((secx-tanx)(sec+tanx))dx`

`rArrint((secx+tanx)secxtanx)/(sec^2x-tan^2x)dx`

as, `color(magenta)(sec^2x-tan^2x=1`

`rArrint(secx+tanx)xxsecxtanxdx`

Let, `color(red)(secx=t`, therefore ,`secxtanxdx=dt`

`rArrI=int(t+sqrt(t^2-1))dt`

`=intt.dt+intsqrt(t^2-1).dt`

Applying, `color(green)(intsqrt(X^2-a^2)dX=X/2sqrt(X^2-a^2)-a^2/2*ln|X+sqrt(X^2-a^2)|+C`

`=t^2/2+[t/2sqrt(t^2-1)-1/2*ln(t+sqrt(t^2-1))]+C`

Replacing , `t=secx`

`rArrI=sec^2x/2+[secx/2sqrt(sec^2x-1)-1/2*ln|secx+sqrt(sec^2x-1|)]+C`

`rArrI=sec^2x/2[1/2secxcolor(magenta)(tanx)-1/2*ln |secx+color(magenta)(tanx)|]+C`

We need

`cos^2x=1-sin^2x`

Therefore,

`int(tanxdx)/(1-sinx)=int(sinxdx)/(cosx(1-sinx))`

`=int(sinxcosxdx)/(cos^2x(1-sinx))`

`=int(sinxcosxdx)/((1-sin^2x)(1-sinx))`

`=int(sinxcosxdx)/((1+sinx)(1-sinx)^2)`

Perform the substitution

`u=sinx`, `=>`, `du=cosxdx`

Therefore,

`int(tanxdx)/(1-sinx)=int(udu)/((1+u)(1-u)^2)`

Perform the decomposition into partial fractions

`u/((1+u)(1-u)^2)=A/(1+u)+B/(1-u)^2+C/(1-u)`

`=(A(1-u)^2+B(1+u)+C(1+u)(1-u))/(((1+u)(1-u)^2))`

The denominators are the same, compare the numerators

`u=A(1-u)^2+B(1+u)+C(1+u)(1-u)`

Let `u=-1`, `=>`, `-1=4A`, `=>`, `A=-1/4`

Let `u=1`, `=>`, `1=2B`, `=>`, `B=1/2`

Coefficients of `u^2`

`0=A-C`, `=>`, `C=A=-1/4`

Therefore,

`u/((1+u)(1-u)^2)=(-1/4)/(1+u)+(1/2)/(1-u)^2+(-1/4)/(1-u)`

So,

`int(tanxdx)/(1-sinx)=int(-1/4du)/(1+u)+int(1/2du)/(1-u)^2+int(-1/4du)/(1-u)`

`=-1/4ln(1+u)-1/2*1/(1-u)+1/4ln(1-u)`

`=-1/4ln(1+sinx)-1/(2(1-sinx))+1/4ln(1-sinx)+C`