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I'm having a lot, I repeat, a lot of trouble with Calculus II, particularly trigonometric substitution. At the moment, I'm extremely confused as to how to integrate $\int x\sqrt{1+x^2}\,\mathrm dx$ using trigonometric substitution. Here's what I have so far):

$$\int x\sqrt{1+x^2}\,\mathrm dx, x=\tan\theta, dx=\sec^2\theta d\theta$$

$$\int \tan\theta \sec^3\theta \,\mathrm dx, u=\sec\theta, \,\mathrm du=\tan\theta \sec\theta \,\mathrm d\theta$$

$$\int u^2 \,\mathrm du$$

$$\frac13\tan\theta + C$$

$$\frac13\sqrt{1+x^2} + C$$

However, the book says the answer is $\frac13(1+x^2)^{3/2} + C$. I don't understand where that extra magnitude of power coming from. Where did I mess up, and what strategies can I use to prevent me from making that and similar mistakes in the future? Thank you in advance for your responses.

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    $\begingroup$ Please use Latex for better legibility. $\endgroup$ – Sherlock Holmes Oct 19 '14 at 5:58
  • $\begingroup$ $x\,{\rm d}x = {1 \over 2}\,{\rm d}\left(x^{2}\right)$. $\endgroup$ – Felix Marin Oct 19 '14 at 22:51
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When you integrate $\int w^2 dw$, you should get $\frac{w^3}{3}+c=\frac{\sec^3x}{3}+c=\frac{(1+x^2)^{\frac{3}{2}}}{3}+c$

Your error lies in the fact that despite $u=\sec \theta$, you suddenly just let the integral be the primitive of $\frac{\tan \theta}{3}+c$.

To prevent making similar mistakes in the future, just make sure that you're consistent with your algebra and read each line carefully (sorry if this isn't of much assistance but it's all I can really think of).

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  • $\begingroup$ First of all, thank you for letting me know what the syntax language is called. I was scratching my head wondering how to format my post correctly. Anyway, I get exactly what you're saying. Even though the integral of $sec^2\theta$ is $tan\theta$, I cannot do that because I used u substitution, correct? Unfortunately, I have and A=1, O=x, and H=$\sqrt{1+x^2}$, therefore $sec\theta$ as 1/1. I must be mistaken on which variables belong to which side of the triangle, but how? $\endgroup$ – Todd Oct 19 '14 at 6:08
  • $\begingroup$ @Todd: There is a MathJax tutorial here. $\endgroup$ – robjohn Oct 19 '14 at 6:17
  • $\begingroup$ @robjohn: Thank you as well! Fractions were eluding me. $\endgroup$ – Todd Oct 19 '14 at 6:22
  • $\begingroup$ Oh, wow. It's far too late for me to be doing homework - I was actually thinking secant was 1/A, not H/A. It's silly mistakes like that that always get me. sigh thank you for your help, everyone. $\endgroup$ – Todd Oct 19 '14 at 6:41
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We don't need to use trig substitutions in this problem. $$ \begin{align} \int x\sqrt{1+x^2}\,\mathrm{d}x &=\frac12\int(1+x^2)^{1/2}\,\mathrm{d}(1+x^2)\\ &=\frac12\left(\frac23(1+x^2)^{3/2}+2C\right)\\ &=\frac13(1+x^2)^{3/2}+C \end{align} $$


Note that in your answer you have $$ \int u^2\,\mathrm{d}u=\frac13\tan(\theta)+C $$ where it should be $$ \begin{align} \int u^2\,\mathrm{d}u &=\frac13u^3+C\\ &=\frac13\sec^3(\theta)+C \end{align} $$ which will give the correct answer since $\sec(\theta)=\sqrt{1+x^2}$.

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  • $\begingroup$ I fully realize that trig substitutions are not necessary to use this problem, and believe me, I wish that my textbook did not require me to use them it. Thank you for the answer though! $\endgroup$ – Todd Oct 19 '14 at 6:23
  • $\begingroup$ Okay. I like to avoid using trig substitutions if they can easily be seen not to be needed. In that case, Sherlock's answer fixing the integration mistake is what you need. $\endgroup$ – robjohn Oct 19 '14 at 6:24
  • $\begingroup$ My post below confirms this because $$\int \tan\left(\theta\right) \sec^3\left(\theta\right)d\theta=\frac{\sec^3\left(\theta\right)}{3}+C_1.$$ $\endgroup$ – jm324354 Oct 19 '14 at 6:35
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You're right by using the substitution $x=\tan\left(\theta\right)$, so your problem becomes

$$\int x\sqrt{1+x^2} dx,$$

\begin{equation}\text{since we may substitute}\: \left\{\begin{array}{lr} x=\tan\left(\theta\right),\:\:dx=\sec^{2}\left(\theta\right)d\theta, \\ 1+x^2=1+\tan^2\left(\theta\right)=\sec^{2}\left(\theta\right). \end{array}\right. \end{equation}

\begin{equation} \therefore \int x\sqrt{1+x^2}dx=\int\tan\left(\theta\right)\cdot\sqrt{\sec^2\left(\theta\right)}\:sec^2\left(\theta\right)\:d\theta=\int\tan\left(\theta\right)\sec^3\left(\theta\right)d\theta=\left(\frac{1}{3}\right)\sec^{3}\left(\theta\right), \end{equation} and now we re-substitute in terms of $x$ and find that \begin{equation} \left(\frac{1}{3}\right)\sec^3\left(\theta\right)=\left(\frac{1}{3}\right)\left(\sqrt{1+x^2}\right)^{3}=\boxed{\frac{\left(1+x^2\right)^\frac{3}{2}}{3}+C_1.} \end{equation}

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  • $\begingroup$ I wish I could give the check-mark to two people, but Sherlock answered first and his answer was also correct. Regardless, I appreciate your post and the clarification. $\endgroup$ – Todd Oct 19 '14 at 6:42

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