How do you integrate #int x^3sqrt(x^2-1)# using substitution?

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Answer 1 · 2016-07-29T13:45:48

#1/15(3x^2+2)(x^2-1)^(3/2)+C#.

Let us take the subst. #x^2-1=t^2#, so, #x^2=t^2+1#, &,

#2xdx=2tdt, or, xdx=tdt#

Now, #I=intx^3sqrt(x^2-1)dx=intx^2sqrt(x^2-1)xdx#

#=int(t^2+1)(sqrt(t^2))tdt=int(t^4+t^2)dt=t^5/5+t^3/3#

#=t^3/15(3t^2+5)=t/15*t^2(3t^2+5)#

#=sqrt(x^2-1)/15*(x^2-1){3(x^2-1)+5}#

#=1/15(3x^2+2)(x^2-1)^(3/2)+C#.

Answer 2 · 2016-07-29T15:47:37

Here is a third solution.

#intx^3sqrt(x^2-1) dx = int x^2sqrt(x^2-1) x dx#

Let #u = x^2-1#,

so that #du = 2xdx#

and #x^2 = u+1#.

The integral becomes

#int (u+1)u^(1/2) 1/2 du = 1/2 int (u^(3/2)+u^(1/2))du#

# = 1/2[2/5u^(5/2)+2/3u^(1/2)] +C#

# = 1/15[3u^(5/2)+5u^(3/2)]+C#

# = 1/15 u^(3/2)[3u+5]+C#

Back-substituting gets us

# = 1/15 (x^2-1)^(3/2)[3(x^2-1)+5]+C#

# = 1/15 (x^2-1)^(3/2)(3x^2+2)+C#