<s>[INST] 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
 [/INST] 
By the above diagram, we can show that
$$
\begin{aligned}
I &=6(\tan \theta \sec \theta+\ln |\sec \theta+\tan \theta|)+c \\
&=6\left(\frac{x}{\sqrt{2}} \cdot \frac{\sqrt{x^{2}+2}}{\sqrt{2}}+\ln \left|\frac{x}{\sqrt{2}}+\frac{\sqrt{x^{2}+2}}{\sqrt{2}}\right|\right)+c \\
&=3x \sqrt{x^{2}+2}+6 \ln \left|x+\sqrt{x^{2}+2}\right|+C
\end{aligned}
$$
:|D Wish it can help!
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