Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 16

int_(sqrt(2)/3)^(2/3)1/(x^5sqrt(9x^2-1))dx
Let's first evaluate the indefinite integral by integral substitution,
Let x=1/3sec(u)
=>dx=1/3sec(u)tan(u)du
int1/(x^5sqrt(9x^2-1))dx=int(1/((1/3sec(u))^5sqrt(9(1/3sec(u))^2-1)))1/3sec(u)tan(u)du
=int(1/(1/243sec^5(u)sqrt(sec^2(u)-1)))1/3sec(u)tan(u)du
Now use the identity:sec^2(theta)=1+tan^2(theta)
=int(243/(3sec^5(u)sqrt(1+tan^2(u)-1)))sec(u)tan(u)du
=int(81sec(u)tan(u))/(sec^5(u)sqrt(tan^2(u)))du
=81int1/(sec^4(u))du
=81intcos^4(u)du
Now let's use the identity:cos^2(theta)=(1+cos(2theta))/2
=81int((1+cos(2u))/2)^2du
=81int(1+cos^2(2u)+2cos(2u))/4du
=81int(1/4+(cos^2(2u))/4+1/2cos(2u))du
=81(int1/4du+int(cos^2(2u))/4du+int(cos(2u))/2du)
=81(u/4+1/4int(1+cos(4u))/2du+1/2intcos(2u)du)
=81(u/4+1/4int(1/2+cos(4u)/2)du+1/2(sin(2u))/2)
=81(u/4+1/4(int1/2du+intcos(4u)/2du)+1/4sin(2u))
=81(u/4+1/4(u/2+1/2sin(4u)/4)+1/4sin(2u)) =81(u/4+u/8+sin(4u)/32+sin(2u)/4)
=81((3u)/8+sin(4u)/32+sin(2u)/4)
Now recall that we have used x=1/3sec(u)
=>sec(u)=3x
=>cos(u)=1/(3x)
=>u=arccos(1/(3x))
Substitute back u and add a constant C to the solution,
=81(3/8arccos(1/(3x))+1/32sin(4arccos(1/(3x)))+1/4sin(2arccos(1/(3x))))+C
Now let's evaluate the definite integral,
int_(sqrt(2)/3)^(2/3)dx/(x^5sqrt(9x^2-1))=81[3/8arccos(1/(3x))+1/32sin(4arccos(1/(3x)))+1/4sin(2arccos(1/(3x)))]_(sqrt(2)/3)^(2/3)
=81[3/8arccos(1/2)+1/32sin(4arccos(1/2))+1/4sin(2arccos(1/2))]-81[3/8arccos(1/sqrt(2))+1/32sin(4arccos(1/sqrt(2)))+1/4sin(2arccos(1/sqrt(2)))]
=81[3/8*pi/3+1/32sin(4*pi/3)+1/4sin(2*pi/3)]-81[3/8*pi/4+1/32sin(4*pi/4)+1/4sin(2*pi/4)]
=81[pi/8+1/32sin((4pi)/3)+1/4sin((2pi)/3)]-81[(3pi)/32+1/32sin(pi)+1/4sin(pi/2)]
=81[pi/8+1/32(-sqrt(3)/2)+1/4(sqrt(3)/2)]-81[(3pi)/32+1/32(0)+1/4(1)]
=81[pi/8+sqrt(3)/2(-1/32+1/4)-(3pi)/32-1/4]
=81[pi/8-(3pi)/32+sqrt(3)/2(7/32)-1/4]
=81[pi/32+(7sqrt(3))/64-1/4]