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. 

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*}
