How do you find $\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x$ $int x^3/sqrt(2x^2 + 4) dx$ using trigonometric substitution?

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How do you find $\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x$ $int x^3/sqrt(2x^2 + 4) dx$ using trigonometric substitution?

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Answer 1 · 2018-06-04T09:36:47

$\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x = \frac{(x^{2} - 4) \sqrt{2 x^{2} + 4}}{6} + C$ $int x^3/sqrt(2x^2+4)dx = ((x^2 -4) sqrt(2x^2+4))/6+C$

Explanation:

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

$\frac{d}{d x} (\sqrt{2 x^{2} + 4}) = \frac{2 x}{\sqrt{2 x^{2} + 4}}$ $d/dx (sqrt(2x^2+4)) = (2x)/sqrt(2x^2+4)$

we can integrate by parts:

$\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x = \int \frac{x^{2}}{2} \frac{d}{d x} (\sqrt{2 x^{2} + 4}) d x$ $int x^3/sqrt(2x^2+4)dx = int x^2/2 d/dx (sqrt(2x^2+4))dx$

$\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x = \frac{x^{2} \sqrt{2 x^{2} + 4}}{2} - \int \frac{d}{d x} (\frac{x^{2}}{2}) \sqrt{2 x^{2} + 4} d x$ $int x^3/sqrt(2x^2+4)dx = ( x^2sqrt(2x^2+4))/2 - int d/dx(x^2/2) sqrt(2x^2+4)dx$

$\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x = \frac{x^{2} \sqrt{2 x^{2} + 4}}{2} - \int x \sqrt{2 x^{2} + 4} d x$ $int x^3/sqrt(2x^2+4)dx = ( x^2sqrt(2x^2+4))/2 - int xsqrt(2x^2+4)dx$

Solve the resulting integral by substitution:

$t = 2 x^{2} + 4$ $t = 2x^2+4$

$d t = 4 x d x$ $dt = 4xdx$

$\int x \sqrt{2 x^{2} + 4} d x = \frac{1}{4} \int \sqrt{t} d t$ $int xsqrt(2x^2+4)dx = 1/4 int sqrt t dt$

$\int x \sqrt{2 x^{2} + 4} d x = \frac{1}{4} \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{1}{6} t \sqrt{t} + C$ $int xsqrt(2x^2+4)dx = 1/4 t^(3/2)/(3/2)+ C = 1/6 t sqrt t +C$

and undoing the substitution:

$\int x \sqrt{2 x^{2} + 4} d x = \frac{x^{2} + 2}{3} \sqrt{2 x^{2} + 4} + C$ $int xsqrt(2x^2+4)dx = (x^2+2)/3 sqrt(2x^2+4)+C$

Putting the results together:

$\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x = \frac{x^{2} \sqrt{2 x^{2} + 4}}{2} - \frac{x^{2} + 2}{3} \sqrt{2 x^{2} + 4} + C$ $int x^3/sqrt(2x^2+4)dx = ( x^2sqrt(2x^2+4))/2 - (x^2+2)/3 sqrt(2x^2+4)+C$

and simplifying:

$\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x = (\frac{x^{2}}{2} - \frac{x^{2} + 2}{3}) \sqrt{2 x^{2} + 4} + C$ $int x^3/sqrt(2x^2+4)dx = (( x^2)/2 - (x^2+2)/3) sqrt(2x^2+4)+C$

$\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x = (3 x^{2} - 2 x^{2} - 4) \frac{\sqrt{2 x^{2} + 4}}{6} + C$ $int x^3/sqrt(2x^2+4)dx = (3x^2 - 2x^2-4) sqrt(2x^2+4)/6+C$

$\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x = \frac{(x^{2} - 4) \sqrt{2 x^{2} + 4}}{6} + C$ $int x^3/sqrt(2x^2+4)dx = ((x^2 -4) sqrt(2x^2+4))/6+C$

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How do you find $\int \frac{x^{3}}{\sqrt{2 x^{2} + 4}} d x$ $int x^3/sqrt(2x^2 + 4) dx$ using trigonometric substitution?

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