How do you integrate #int 1/(x^2sqrt(16x^2-9))# by trigonometric substitution?

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Answer 1 · 2018-03-03T18:22:45

# sqrt(16x^2-9)/(9x)+C#.

Suppose that, #I=int1/(x^2sqrt(16x^2-9))dx=int1/(x^2sqrt{(4x)^2-3^2})dx#.

We subst. #4x=3secy," so that, "4dx=3secytanydy#.

#:. I=int1/{(3/4*secy)^2sqrt(9sec^2y-9)}*(3/4*secytany)dy#,

#=16/9*1/4int1/(sec^2ytany)*secytanydy#,

#=4/9int1/secydy#,

#=4/9intcosydy#,

#=4/9siny#,

#=4/9sqrt(1-cos^2y)#,

#=4/9sqrt(1-1/sec^2y)#,

#=4/9sqrt{1-1/(4/3*x)^2}#,

#=4/9sqrt{1-9/(16x^2)}#.

# rArr I=sqrt(16x^2-9)/(9x)+C#.