#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#