What is the integral of x^3 over square rood of 1+x^2 dx by using the substitution u=sinh (x)?

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Mar 8, 2018

The answer is $= \frac{1}{3} \left({x}^{2} - 2\right) \sqrt{{x}^{2} + 1} + C$

Explanation:

We need

${\cosh}^{2} x - {\sinh}^{2} x = 1$

$\arcsin h x = \ln \left(x + \sqrt{{x}^{2} + 1}\right)$

I prefer to use the substitution

$x = \sinh u$, $\implies$, $\mathrm{dx} = \cosh u \mathrm{du}$

Therefore,

$\int \frac{{x}^{3} \mathrm{dx}}{\sqrt{1 + {x}^{2}}} = \int \frac{{\sinh}^{3} u \cosh u \mathrm{du}}{\cosh u}$

$= \int {\sinh}^{3} u \mathrm{du}$

$= \int {\sinh}^{2} u \sinh u \mathrm{du}$

$= \int \left({\cosh}^{2} u - 1\right) \sinh u \mathrm{du}$

Let $v = \cosh u$, $\implies$, $\mathrm{dv} = \sinh u \mathrm{du}$

Therefore,

$\int \left({\cosh}^{2} u - 1\right) \sinh u \mathrm{du} = \int \left({v}^{2} - 1\right) \mathrm{dv}$

$= \frac{1}{3} {v}^{3} - v$

$= \frac{1}{3} {\cosh}^{3} u - \cosh u$

$= \frac{1}{3} {\cosh}^{3} \left(\arcsin h x\right) - \cosh \arcsin h x + C$

$\cosh \left(\arcsin h \left(x\right)\right) = \frac{1}{2} \left({e}^{\arcsin h x} + {e}^{- \arcsin h x}\right)$

$= \frac{1}{2} \left({e}^{\ln \left(x + \sqrt{{x}^{2} + 1}\right)} + {e}^{- \ln \left(x + \sqrt{{x}^{2} + 1}\right)}\right)$

$= \frac{1}{2} \left(\left(x + \sqrt{{x}^{2} + 1}\right) + \frac{1}{x + \sqrt{{x}^{2} + 1}}\right)$

$= \frac{1}{2} \left(\left(x + \sqrt{{x}^{2} + 1}\right) + \frac{x - \sqrt{{x}^{2} + 1}}{x + \sqrt{{x}^{2} + 1} \left(x - \sqrt{{x}^{2} + 1}\right)}\right)$

$= \frac{1}{2} \left(\left(x + \sqrt{{x}^{2} + 1}\right) + \left(x - \frac{\sqrt{{x}^{2} - 1}}{{x}^{2} - {x}^{2} - 1}\right)\right)$

$= \sqrt{{x}^{2} + 1}$

And

${\cosh}^{3} \left(\arcsin h x\right) = {\left(\sqrt{{x}^{2} + 1}\right)}^{3}$

$= \left({x}^{2} + 1\right) \sqrt{{x}^{2} + 1}$

Finally,

$\int \frac{{x}^{3} \mathrm{dx}}{\sqrt{1 + {x}^{2}}} = \frac{1}{3} \left({x}^{2} + 1\right) \sqrt{{x}^{2} + 1} - \sqrt{{x}^{2} + 1} + C$

$= \frac{1}{3} \left({x}^{2} - 2\right) \sqrt{{x}^{2} + 1} + C$

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