Calculus 2 - $\int(\sqrt{72+36x^2}dx$ I have done this problem several times and this is the only answer i ever come to. My schools webwork gives me incorrect for my answer (answer is not simplified but it should be accepted in this format). Did i do this correctly?
Here is my work:
\begin{align}
\int \sqrt{72+36x^2}\, dx&=\sqrt {36}\int \sqrt{2+x^2}\,dx\\
&=6\int \sqrt{2+x^2}\,dx\\
&=6\int \sqrt 2 \sec \theta \sqrt 2 \sec^2 \theta \, d\theta\\
&=12 \int sec^3 \theta=12\left[\frac{\tan \theta \sec \theta}2 +\frac 12 \int \sec\theta \, d\theta\right]\\
&=12\left[\frac{\tan\theta\sec\theta}2+\frac 12 \ln|\sec\theta+\tan\theta|\right]+C\\
&=6\tan \theta\sec\theta+6\ln|\sec\theta+\tan\theta|+C\\
&=6\tan\left(\tan^{-1}\frac{x}{\sqrt 2}\right)\sec\left(\tan^{-1}\frac{x}{\sqrt 2}\right)\\
&+6\ln \left|\sec\left(\tan^{-1}\frac{x}2\right)+\tan\left(\tan^{-1}\frac{x}{\sqrt 2}\right)\right|+C
\end{align}
Any help is appreciated. Thanks

I believe that your work is correct. However, at the end you are most likely asked to put this in a nicer form. The way to simplify $trig_1(trig_2^{-1}(\frac{a}{b}))$ is form a right triangle that fits your $trig^{-1}$ conditions and then compute $trig_1(angle)$.
As an example, simplifying $\tan{(\arctan{\frac{x}{\sqrt{2}}})}$ means forming a right triangle where the opposite side from an angle (we'll call $\theta$) is $x$ and the adjacent side from $\theta$ is $\sqrt{2}$. The hypotenuse of this triangle is obviously then $\sqrt{2+x^2}$. Therefore, 
$$\tan{(\arctan{\frac{x}{\sqrt{2}}})}=\tan{\theta} = \frac{x}{\sqrt{2}}$$
The above could also be seen by recognizing that tangent and arctangent are inverse functions. Using this same triangle, we also have: 
$$\sec{(\arctan{\frac{x}{\sqrt{2}}})} = \sec{\theta} = \frac{\sqrt{2+x^2}}{\sqrt{2}}$$
I'll leave you to simplify the rest.
