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Today I had to take the indefinite integral $\int x^3 \sqrt{x^2-1} \, dx$

My steps:

  1. $x=\sec\theta$, $dx=\sec\theta\tan\theta\, d\theta$
  2. $\displaystyle \int \sec^3\theta \sqrt{\sec^2\theta - 1} \, \sec\theta\tan\theta\, d\theta$
  3. ${\displaystyle \int \sec^3\theta \sqrt{\tan^2\theta}\sec\theta\tan\theta d\theta}$
  4. ${\displaystyle \int \sec^3\theta \tan\theta \cdot \sec\theta \tan\theta d\theta}$
  5. ${\displaystyle \int \sec^4\theta \cdot \tan^2\theta d\theta}$
  6. ${\displaystyle \int \sec^2\theta \sec^2\theta\tan^2\theta d\theta}$
  7. $u=\tan\theta$, $du= \sec^2\theta d\theta$
  8. ${\displaystyle \int u^2(u^2 + 1)}$
  9. ${\displaystyle \int u^4 + u^2}$
  10. ${\displaystyle = \frac{1}{5}u^5 + \frac{1}{3}u^3 + C}$

here is where I'm even more shaky:

  1. ${\displaystyle \frac{1}{5}\tan^5 + \frac{1}{3}\tan^3 + C}$
  2. since $\tan= \sin/\cos$, $\tan$ also $= \sec/\csc$, and if $\sec\theta=x$, per the substitution, couldn't $\tan$ also be written $x/(1/x)=x^2$, right?
  3. I replaced all $\tan$s in the final answer with $x^2$, leaving a final answer of ${\displaystyle \frac{1}{5}x^{10}+\frac{1}{3}x^6+C}$
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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig May 3 '16 at 12:09
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You are good except for your last step; $\tan(\theta)$ is not $x^2$, and in particular $\csc$ is not $1/\sec$.

One way to identify what $\tan(\theta)$ is in terms of $x$ is to consider the triangle that justifies the substitution $x=\sec(\theta)$ in the first place. This has a leg of $\sqrt{x^2-1}$, which means the hypotenuse is $x$ and the other leg is $1$. By choosing $u=\sec(\theta)$ you make $\theta$ adjacent to the leg of length $1$ and opposite the leg of length $\sqrt{x^2-1}$. Thus $\tan(\theta)=\sqrt{x^2-1}$. Alternately you could have figured that out by simply using the identity $\sec^2(\theta)=\tan^2(\theta)+1$.

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  • $\begingroup$ was there any way to avoid using a triangle? $\endgroup$ – KrissyMichaelsson May 2 '16 at 15:20
  • $\begingroup$ @KrissyMichaelsson Yes, see my last sentence. Still, I personally like using a triangle a lot, even if I only just imagine it rather than actually drawing it. $\endgroup$ – Ian May 2 '16 at 15:20
  • $\begingroup$ Ohhh I see my mistake now....if x=secΘ, I should said x=sqrt(tan^2+1, and then x^2=tan^2+1, x^2-1=tan^2, then sqrt(x^2-1)=tan. Thanks. I was in a test, so it was def a rush, since I knew triangles are the best way to do it, but I forgot how to do that. $\endgroup$ – KrissyMichaelsson May 2 '16 at 15:32
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How about this: \begin{align} u & = \sqrt{x^2-1} \\ u^2 & = x^2 - 1 \\ u^2+1 & = x^2 \\ 2u\,du & = 2x\,dx \\ \end{align} \begin{align} \int x^3 \sqrt{x^2-1} \, dx & = \int x^2 \sqrt{x^2-1} \Big( x\,dx\Big) \\ & = \int (u^2+1) u \left( \frac 1 2 \, du \right) \\ & = \frac 1 2 \int (u^4 + u^2) \, du \\ & \phantom{{}={}} \text{etc.} \end{align}

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  • $\begingroup$ we were required to use trig substitution, tho that would've been a great way of checking. $\endgroup$ – KrissyMichaelsson May 2 '16 at 21:28
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This would have been lots easier with $x=\cosh\theta$: $$\begin{align}\int x^3\sqrt{x^2-1}dx&=\int\cosh^3\theta\sinh^2\theta\,d\theta\\ &=\int(\sinh^2\theta+1)\sinh^2\theta\cosh\theta\,d\theta\\ &=\frac15\sinh^5\theta+\frac13\sinh^3\theta+C\\ &=\frac15(x^2-1)^{5/2}+\frac13(x^2-1)^{3/2}+C\\ &=\left(\frac15x^2-\frac15+\frac13\right)(x^2-1)^{3/2}+C\\ &=\frac1{15}(3x^2+2)(x^2-1)^{3/2}+C\end{align}$$ But I don't know if hyperbolic substitution counts as trigonometric substitution in this context.

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