Start by multiplying both numerator and denominator by cos(x)
#intcos(x)/(cos^2(x)*(1-sin(x))^2)dx#
Let's #t = sin(x)#
#dt = cos(x) dx#
Remember pythagore #cos^2(x)=1-sin^2(x)#
Now we have
#=>int1/((1-t^2)*(1-t)^2)dt#
Factor the denominator :
#=>int-1/((t-1)^3*(t+1))dt #
Now partial fraction !
#=>-1/((t-1)^3*(t+1)) =#
#-theta/(t-1)-beta/(t-1)^2-alpha/(t-1)^3-nu/(t+1)#
Do some boring math :
Multiply both side by #-(t-1)^3*(t+1)# and simplify
You will find :
#=>1=(theta+nu)t^3+(-theta+beta-3nu)t^2+(-theta+alpha+3nu)t+theta-beta+alpha-nu#
4 systems 4 unknown
Result is :
#theta = 1/8#
#beta = -1/4#
#alpha = 1/2#
#nu = -1/8#
#=>int1/((1-t^2)*(1-t)^2)dt# =
#=>-1/8int1/(t-1)+1/4int1/(t-1)^2-1/2int1/(t-1)^3+1/8int1/(t+1)#
#=>-1/8[ln(|t-1|)]-1/4[1/(t-1)]+1/4[1/(t-1)^2]+1/8[ln(|t+1|)]#
Substitute back for #t = sin(x)#
#=>-1/8[ln(|sin(x)-1|)]-1/4[1/(sin(x)-1)]+1/4[1/(sin(x)-1)^2]+1/8[ln(|sin(x)+1|)]#
