Integrate 1/((x^5)sqrt(9x^2-1)) ...?

1 answer:

5 0

Let 9x^2-1 = y^2
=> 18xdx = 2ydy 
=> ydy = 9xdx 

lower limit = sqrt(9*2/9 - 1) = sqrt(1) = 1 
upper limit = sqrt(9*4/9 - 1) = sqrt(3) 

Int. [sqrt(2)/3,2/3] 1/(x^5(sqrt(9x^2-1)) dx 

= Int. [sqrt(2)/3,2/3] xdx/(x^6(sqrt(9x^2-1)) 
= 81* Int. [1,sqrt(3)] ydy/((y^2+1)^3y) 
=81* Int. [1,sqrt(3)] dy/(y^2+1)^3 

y=tanz 
dy = sec^2z dz 

=81*Int [pi/4,pi/3] cos^4(z) dz 
=81/4*int [pi/4,pi/3] (1+cos(2z))^2 dz 
=81/4* Int. [pi/4,pi/3] (1+2cos(2z)+cos^2(2z)) dz 

=81/4*(pi/3-pi/4) + 81/4*(sin(2pi/3)-sin(pi/2)) + 81/8 * (pi/3-pi/4) 
+ 81/32 *(sin(-pi/3)-sin(pi)) 
=81(pi/48+pi/96+1/4*(sqrt(3)/2 - 1) - 1/32 * sqrt(3)/2) 
=81/32*(pi+3sqrt(3)-8)

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Please find the attached diagram for a better understanding of the question.

As we can see from the diagram,

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$\angle x=\angle RPQ$ = which is unknown.

Let us begin solving now. The first step is to find the angle $\angle x$ which can be found by using the following trigonometric ratio in $\Delta PQR$ :