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|)]
