Evaluate the integral with hyperbolic or trigonometric substitution. ?

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1 Answer
Yahia M.
Mar 29, 2018

#sqrt(6)/144sin(2sec^-1sqrt(6)x/6)+V/72sec^-1(sqrt(6)x/6)+C#

Explanation:

by applying the trigonometric substitution:
#x=sqrt(6)sec(u)#
#dx=sqrt(6)sec(u)*tan(u)du#
#int((sqrt(6)secu*tanudu)/(6*sqrt(6)*sec^2u*sqrt(6)tanu))#

after simplification
#sqrt(6)/36int(1/sec^2(u)du)#=#sqrt(6)/36int(cos^2(u)du)#

by using double-angle formulae: #cos^2u=1/2(1+cos2u)#

#sqrt(6)/72int(1+cos2u)du#

=#sqrt(6)/72(u+1/2sin2u)#
by substituting back #u=sec^-1(x/sqrt(6))#
you get:
#intdx/(x^3*sqrt(x^2-6))=sqrt(6)/144sin(2sec^-1sqrt(6)x/6)+V/72sec^-1(sqrt(6)x/6)+C#

I hope this was helpful

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