Inverse Trig & Trig Sub

Question

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.

Answer 1

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*}

Answer 2

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

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

Answer 3

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 $\ u=1+x^2$

Answer 4

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

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