# Inverse Trig & Trig Sub

Can someone explain to me how to solve this using inverse trig and trig sub?

$$\int\frac{x^3}{\sqrt{1+x^2}}\, dx$$

Thank you.

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You can use hyperbolic substitution, i.e. let $x=\sinh t$ then \begin{align*} \int\frac{x^3}{\sqrt{1+x^2}}\, dx&=\int \sinh^3 t\, dt\\ &=\int (\cosh^2t -1)\sinh t\, dt\\ &=\frac{\cosh^3 t}{3}-\cosh t+C\\ &=\frac{(\sqrt{1+x^2})^3}{3}-\sqrt{1+x^2}+C \end{align*} Also you can use triangle substitution : $x=\tan \theta$ \begin{align*} \int\frac{x^3}{\sqrt{1+x^2}}\, dx&=\int \tan^3\theta\sec\theta\, d\theta\\ &=\int (\sec^2\theta-1)\sec\theta\tan\theta\, d\theta\\ &=\frac{\sec^3\theta}{3}-\sec\theta+C\\ &=\frac{(\sqrt{1+x^2})^3}{3}-\sqrt{1+x^2}+C \end{align*}

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How do you explain the different results? – egreg Dec 14 '13 at 12:07
I have a mistake I've corrected, thanks for your comment. – Farshad Nahangi Dec 14 '13 at 14:37
How did u know to plug in tan theta? Tan is A^2 + x^2, but there is a square root over that. – user113113 Dec 14 '13 at 18:27
Dear @user113113; Indeed when you have $\sqrt{a^2+x^2}$ it will be useful. Since $a^2+x^2=a^2sec^2\theta$ square root will be omitted. – Farshad Nahangi Dec 14 '13 at 18:51

Set $u=1+x^2$. Then $\dfrac{du}{dx}=2x$ and so,

$\displaystyle\int \dfrac{x^3}{\sqrt{1+x^2}}dx=\int \dfrac12\dfrac{u-1}{\sqrt{u}}du=\dfrac12\int (u^{1/2}-u^{-1/2})\, du$.

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nice observation (+1) – lab bhattacharjee Dec 14 '13 at 11:33
@labbhattacharjee Thank you! – Alraxite Dec 14 '13 at 11:48

$$\frac{x^3}{\sqrt{1+x^2}}=\frac{x^3+x-x}{\sqrt{1+x^2}}=x\sqrt{1+x^2}-\frac x{\sqrt{1+x^2}}$$

Set $1+x^2=u$ in each case

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$\dfrac{x^3+x^2}{\sqrt{1+x^2}}$ doesn't simplify to $x\sqrt{1+x^2}$. – Alraxite Dec 14 '13 at 11:49
@Alraxite, thanks for your observation. Wrong calculation made things unnecessarily complex. – lab bhattacharjee Dec 14 '13 at 11:53

You can also use integration by part: let $u=x^2$ and $dv=\frac{x}{\sqrt{1+x^2}}$ then you will have: \begin{align*} \int udv&=uv-\int vdu\\ &=x^2\sqrt{1+x^2}-\int2x\sqrt{1+x^2}\,dx\\ &=x^2\sqrt{1+x^2}-\frac{2}{3}(1+x^2)^{\frac{3}{2}}+C \end{align*} where the last integral was solved by substitution $\ u=1+x^2$

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$$v=\sqrt{1+x^2}$$ – Ron Gordon Dec 14 '13 at 11:16
Your comment is appreciated @Ron Gordon. – Farshad Nahangi Dec 14 '13 at 11:18