How do you find#int x^3/sqrt(2x^2 + 4) dx # using trigonometric substitution?

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Answer 1 · 2018-06-04T09:36:47

#int x^3/sqrt(2x^2+4)dx = ((x^2 -4) sqrt(2x^2+4))/6+C#

You do not really need to use trigonometric substitutions.
As:

#d/dx (sqrt(2x^2+4)) = (2x)/sqrt(2x^2+4)#

we can integrate by parts:

#int x^3/sqrt(2x^2+4)dx = int x^2/2 d/dx (sqrt(2x^2+4))dx#

#int x^3/sqrt(2x^2+4)dx = ( x^2sqrt(2x^2+4))/2 - int d/dx(x^2/2) sqrt(2x^2+4)dx#

#int x^3/sqrt(2x^2+4)dx = ( x^2sqrt(2x^2+4))/2 - int xsqrt(2x^2+4)dx#

Solve the resulting integral by substitution:

#t = 2x^2+4#

#dt = 4xdx#

#int xsqrt(2x^2+4)dx = 1/4 int sqrt t dt#

#int xsqrt(2x^2+4)dx = 1/4 t^(3/2)/(3/2)+ C = 1/6 t sqrt t +C#

and undoing the substitution:

#int xsqrt(2x^2+4)dx = (x^2+2)/3 sqrt(2x^2+4)+C#

Putting the results together:

#int x^3/sqrt(2x^2+4)dx = ( x^2sqrt(2x^2+4))/2 - (x^2+2)/3 sqrt(2x^2+4)+C#

and simplifying:

#int x^3/sqrt(2x^2+4)dx = (( x^2)/2 - (x^2+2)/3) sqrt(2x^2+4)+C#

#int x^3/sqrt(2x^2+4)dx = (3x^2 - 2x^2-4) sqrt(2x^2+4)/6+C#

#int x^3/sqrt(2x^2+4)dx = ((x^2 -4) sqrt(2x^2+4))/6+C#