How do you evaluate int cot^5(2x) dx?
1 Answer
Apr 19, 2018
Explanation:
We want to integrate
I=intcot^5(2x)dx
Make a substitution
I=1/2intcot^5(u)du
color(white)(I)=1/2int((cos^2(u))^2cos(u))/sin^5(u)du
color(white)(I)=1/2int((1-sin^2(u))^2cos(u))/sin^5(u)du
Make a substitution
I=1/2int(1-s^2)^2/s^5ds
color(white)(I)=1/2int1/s+1/s^5-2 1/s^3ds
color(white)(I)=1/2(ln(s)-1/(4s^4)+1/s^2)+C
color(white)(I)=1/8(4ln(s)-1/s^4+4/s^2)+C
Substitute back
I=1/8(4ln(sin(2x))-csc^4(2x)+4csc^2(2x))+C