Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 28

int5cos(x)/(sin^2(x)+3sin(x)-4)dx
Take the constant out,
=5intcos(x)/(sin^2(x)+3sin(x)-4)dx
Now let's apply integral substitution:u=sin(x)
=>du=cos(x)dx
=5int1/(u^2+3u-4)du
Now to use partial fractions, denominator of the integrand needs to be factored,
Let's split the middle term,
1/(u^2+3u-4)=1/(u^2-u+4u-4)
=1/(u(u-1)+4(u-1))
=1/((u-1)(u+4))
Now let's write it as sum of partial fractions:
1/((u-1)(u+4))=A/(u-1)+B/(u+4)
Multiply the above by the LCD,
=>1=A(u+4)+B(u-1)
1=Au+4A+Bu-B
1=(A+B)u+4A-B
Equating the coefficients of the like terms,
A+B=0 -----------------------------(1)
4A-B=1 ----------------------------(2)
Solve the above linear equations to get the values of A and B,
Add equation 1 and 2,
5A=1
A=1/5
Plug the value of A in equation 1,
1/5+B=0
B=-1/5
Plug in the values of A and B in the partial fraction template,
1/((u-1)(u+4))=(1/5)/(u-1)+(-1/5)/(u+4)
=1/(5(u-1))-1/(5(u+4))
int1/(u^2+3u-4)du=int(1/(5(u-1))-1/(5(u+4)))du
=int1/5(1/(u-1)-1/(u+4))du
Take the constant out,
=1/5int(1/(u-1)-1/(u+4))du
Apply the sum rule,
=1/5(int1/(u-1)du-int1/(u+4)du)
Now use the common integral:int1/xdx=ln|x|
=1/5(ln|u-1|-ln|u+4|)
Substitute back u=sin(x)
=1/5(ln|sin(x)-1|-ln|sin(x)+4|)
int5cos(x)/(sin^2(x)+3sin(x)-4)dx=5(1/5(ln|sin(x)-1|-ln|sin(x)+4|)
Simplify and add a constant C to the solution,
=ln|sin(x)-1|-ln|sin(x)+4|+C