How do you integrate #int (6x^3)/sqrt(9+x^2) dx# using trigonometric substitution?

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Answer 1 · 2018-06-08T11:29:22

# 2(x^2-18)sqrt(x^2+9)+C#.

Suppose that, #I=int(6x^3)/sqrt(9+x^2)dx=int(6x^3)/sqrt(x^2+3^2)dx#.

We subst. #x=3tanu. :. dx=3sec^2udu#.

#:. I=int{(6*27tan^3u)(3sec^2u)}/sqrt(9tan^2u+9)du#,

#=162int(tan^3usecu)#,

#162int(tan^2u*tanusecu)du#.

#=162int(sec^2u-1)secutanudu#.

If we take #secu=v," then, "secutanudu=dv#.

#:. I=162int(v^2-1)dv#,

#=162(v^3/3-v)#,

#=162/3v(v^2-3)#,

#=54secu(sec^2u-3)............[because, v=secu]#,

#=54sqrt(tan^2u+1){(tan^2u+1)-3}#,

#=54{sqrt((x/3)^2+1)}{(x/3)^2-2}#,

# rArr I=2(x^2-18)sqrt(x^2+9)+C#, as desired!

Feel the Joy of Maths.!

Answer 2 · 2018-06-08T11:45:30

Here is a #2^(nd)# Method ( without using trgo. substn.) :

# 2(x^2-18)sqrt(x^2+9)+C#

Let, #I=int(6x^3)/sqrt(x^2+9)dx=int{(3x^2)(2xdx)}/sqrt(x^2+9)#.

Subst. #x^2+9=t^2. :. x^2=t^2-9. :. 2xdx=2tdt#.

#:. I=int{3(t^2-9)(2tdt)}/sqrt(t^2)#,

#=6int(t^2-9)dt#,

#=6(t^3/3-9t)#,

#=2t(t^2-27)#,

#=2{(x^2+9)-27}sqrt(x^2+9)#.

# rArr I=2(x^2-18)sqrt(x^2+9)+C#, as in the #1^(st)# Method!