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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