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

  • 1
    $\begingroup$ 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
  • 4
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.