How do you integrate #int 2/(y^4sqrt(y^2-25)# using trig substitutions?

Our goal here is to get rid of the #√# by making use of a pythagorean identity. Let #y = 5sectheta#. Then #dy = 5secthetatanthetad theta#.

#=int 2/((5sectheta)^4sqrt((5sectheta)^2 - 25)) * 5secthetatanthetad theta#

Apply the identity #sec^2theta - 1 = tan^2theta#:

#=int 2/(625sec^4thetasqrt(25tan^2theta))* 5secthetatantheta d theta#

#=int 2/(625sec^4(5tantheta)) * 5secthetatantheta d theta#

#=int 2/(625sec^3theta) d theta#

Rewrite using #cosx = 1/secx#

#=int 2/625cos^3theta d theta#

#=int 2/625 cos^2theta(costheta) d theta#

Rearrange using the pythagorean identity #cos^2x + sin^2x = 1#:

#=int 2/625 (1 - sin^2theta)costheta d theta#

Let #u = sintheta#. Then #du = costheta d theta# and #d theta = (du)/costheta#.

#=int 2/625 (1 - u^2)costheta * (du)/costheta#

#=int 2/625( 1 - u^2) du#

Integrate using #intx^ndx = x^(n + 1)/(n + 1)# where #n != -1#.

#=2/625(u - 1/3u^3) + C#

#=2/625sintheta - 2/1875sin^3theta + C#

Draw a triangle to represent #y/5 = sectheta#.

#=2/625sqrt(y^2 - 25)/y - 2/1875(sqrt(y^2 - 25)/y)^3 + C#

#=(2sqrt(y^2 - 25))/(625y) - (2(y^2-25)^(3/2))/(1875y^3) + C#

Hopefully this helps!