Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 27

intsin(x)/(cos(x)+cos^2(x))dx
Apply integral substitution: u=cos(x)
=>du=-sin(x)dx
=int1/(u+u^2)(-1)du
Take the constant out,
=-1int1/(u+u^2)du
Now to compute the partial fraction expansion of a proper rational function, we have to factor out the denominator,
=-1int1/(u(u+1))du
Now let's create the partial fraction expansion,
1/(u(u+1))=A/u+B/(u+1)
Multiply the above equation by the denominator,
=>1=A(u+1)+B(u)
1=Au+A+Bu
1=(A+B)u+A
Equating the coefficients of the like terms,
A+B=0 ------------------(1)
A=1
Plug in the value of A in equation 1,
1+B=0
=>B=-1
Plug in the values of A and B in the partial fraction expansion,
1/(u(u+1))=1/u+(-1)/(u+1)
=1/u-1/(u+1)
int1/(u(u+1))du=int(1/u-1/(u+1))du
Apply the sum rule,
=int1/udu-int1/(u+1)du
Now use the common integral:int1/xdx=ln|x|
=ln|u|-ln|u+1|
Substitute back u=cos(x)
=ln|cos(x)|-ln|cos(x)+1|
intsin(x)/(cos(x)+cos^2(x))dx=-1{ln|cos|x|-ln|cos(x)+1|}
Simplify and add a constant C to the solution,
=ln|cos(x)+1|-ln|cos(x)|+C