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Evaluate $\int{x^{2} \sqrt{a^{2} + x^{2}} d x}$ using inverse substitution.

Let $x=a \sinh{\left(u \right)}$

Then $dx=\left(a \sinh{\left(u \right)}\right)^{\prime }du = a \cosh{\left(u \right)} du$

$$\int{x^{2} \sqrt{a^{2} + x^{2}} d x} = \int{a^{2} \sqrt{a^{2} \sinh^{2}{\left( u \right)} + a^{2}} \sinh^{2}{\left( u \right)}a \cosh{\left(u \right)} du}$$

Use the identity: $$\sinh^{2}{\left( u \right)} + 1 = \cosh^{2}{\left( u \right)}$$

Therefore, $$\int{x^{2} \sqrt{a^{2} + x^{2}} d x} = \int{a^{4} \sinh^{2}{\left( u \right)}\cosh^{2}{\left(u \right)} du}$$

Use the double angle formula: $$\sinh{\left( 2u \right)} = 2\sinh{\left( u \right)}\cosh{\left(u \right)}$$

rplee
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1 Answers1

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The next step is to use your identity: $$\sqrt{a^2\sinh^2(u)+a^2}=\sqrt{a^2\left(\sinh^2(u)+1\right)}=\sqrt{a^2\cosh^2(u)}=a\cosh(u) $$ Regarding the signs, $\cosh(x)$ is always positive, and you may take $a\geq 0$. You then get $$a^4\int \sinh^2(u)\cosh^2(u)\,\mathrm{d}u$$ which you integrate the same way you integrate a product of sines and cosines raised to even powers. Use the identities, $\sinh(2x)=2\sinh(x)\cosh(x)$ and $\sinh^2\frac x2=\frac{\cosh(x)-1}{2}$. Once you get rid of the integral sign, you'll get $$\frac{a^4}{8}\left(\frac{\sinh(4u)}{4}-u\right) $$ and you need to convert $u$ to $x$. Note that $u=\sinh^{-1}\left(\frac xa\right)$. As for $\sinh(4u)$, you have, $$\sinh(4u)=2\sinh(2u)\cosh(2u)=4\sinh(u)\cosh(u)\left(2\sinh^2(u)+1\right) $$ using $\cosh(2x)=2\sinh^2(x)+1$. You'll also need $\cosh(\sinh^{-1}(x))=\sqrt{1+x^2}$.

bjorn93
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