Find $\int \sin(2x)\sqrt{1+3\cos^{2}(x)}\ dx$ Is it humanly possible to integrate the following equation?
$$\int \sin(2x)\sqrt{1+3\cos^{2}(x)}\ dx$$
 A: Let $u=\cos^2(x)$, therefore $du=-2\cos(x)\sin(x)\ dx$:
$$\int\sin(2x)\sqrt{1+3\cos^2(x)}\ dx=-\int\sqrt{1+3u}\ du$$
Now, let $v=1+3u$ and $dv=3\ du$:
\begin{align*}
-\int\sqrt{1+3u}\ du&=-\frac13\int\sqrt v\ dv\\
&=-\frac{2v^{3/2}}9+C,\text{ where $C$ is a constant}\\
&=-\frac29(1+3u)^{\frac32}+C\\
&=\boxed{-\frac29(1+3\cos^2(x))^{\frac32}+C}
\end{align*}
A: Hint: Use substitution, and the fact that $\sin 2x = 2\sin x \cos x$.
A: Here's what I'd try, and the reasons why I would want to:
Trig functions with different arguments are BAD!  And there's a simple way to get rid of $\sin(2x)$, so we get:$$\int 2\sin(x)\cos(x)\sqrt{1+3\cos^{2}(x)}\ dx$$with the $2$ easily moved out of the integration.
Next, you've got a function with one $\sin(x)$ and a mess of $\cos(x)$, and $\sin(x)$ is very close to the derivative of $\cos(x)$!  So if we try $$u=\cos(x)$$ $$du=-\sin(x) dx$$ we get $$-2\int u\sqrt{1+3u^2}\ du$$ So now all the trig is gone!
I still have an ugly expression (under a square root, too!), but I also have something close to the differential of the ugly expression.  So I'd try:$$v=1+3u^2$$ $$dv=6u\  du$$ $$u\   du=\frac{dv}{6}$$ giving:$$-\frac{1}{3}\int v^{\frac{1}{2}}\ dv$$This is an integral I can do.  And I'm human... 
