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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

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    $\begingroup$ Your link does not work, for me at least. $\endgroup$ – user228113 Jun 29 '15 at 16:43
  • $\begingroup$ sorry I re-uploaded to imgur $\endgroup$ – ricefieldboy Jun 29 '15 at 16:44
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    $\begingroup$ To avoid downvotes and further issues (e.g. the question being deleted or the link being incomplete) this might help you write mathematics in this site. $\endgroup$ – user228113 Jun 29 '15 at 16:50
  • $\begingroup$ @ricefieldboy I edited your post to make it more readable. Did I make any errors? $\endgroup$ – Cyclohexanol. Jun 29 '15 at 16:58
  • $\begingroup$ thanks alot, there was an error but i fixed it. hopefully someone can answer now $\endgroup$ – ricefieldboy Jun 29 '15 at 17:04
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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.

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