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So I have an integral $\int_1^4\int_y^4\int_0^z\frac{z}{x^2+z^2}\,dx\,dz\,dy$ but I can't figure out what trig substitution to use on the first step. When I try $z=\cos$ and $x=\sin$, I end up with $\int\cos$ but the book comes up with $z\cdot\frac{1}{z}\cdot\arctan \frac{x}{z}$ so I know I screwed up.

Can someone show me how the book took this step?

Thanks!

Michael Hardy
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PoGaMi
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  • Try rewriting the integral with dz as the first integration – zzchan Aug 01 '15 at 22:35
  • My textbook takes the integral in this order, though, so I'm trying to figure out what they did. They have a habit of skipping multiple steps. – PoGaMi Aug 01 '15 at 22:36

1 Answers1

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The innermost integral is with respect to $x$, treating $z$ as a constant.

Since this is the case, you might as well go with: $$ x = z\tan\theta \implies \text{d}x = z \sec^2\theta \text{ d}\theta \qquad (*)$$ $$ x = 0 \implies \theta = 0, \ x = z \implies \theta = \dfrac{\pi}{4} $$

$$ \int_{0}^{z} \dfrac{z}{x^2+z^2} \text{ d}x = \int_{0}^{\pi/4} \dfrac{z}{z^2 \sec^2\theta} \ z\sec^2\theta \text{ d}\theta = \dfrac{\pi}{4} $$

In the case of the indefinite integral, we have that:

$$ \begin{aligned} \int \dfrac{z}{x^2+z^2} \text{ d}x \ \overset{(*)}= \ \int \dfrac{z}{z^2 \sec^2\theta} \ z\sec^2\theta \text{ d}\theta \ = \int z \dfrac{1}{z} \text{ d}\theta \ & = \ z \dfrac{1}{z} \theta + \mathcal{C} \ = \ z \dfrac{1}{z} \arctan \left( \dfrac{x}{z} \right) + \mathcal{C} \end{aligned} $$

Khallil
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  • That's an elegant solution but it doesn't show me how the book arrived at their particular solution. I'm trying to follow their methods in this case, so I'm trying to come out with their same indefinite solution. – PoGaMi Aug 01 '15 at 22:56
  • I think it's exactly the same, but they considered the indefinite integral. – Khallil Aug 01 '15 at 22:57
  • Check out my edit, @PoGaMi. – Khallil Aug 01 '15 at 23:03
  • I worked the indefinite out on paper given your substitution and I get your same answer, which is integral(1), but I don't see where their arctan came from. – PoGaMi Aug 01 '15 at 23:05
  • From our substitution: $$ x=z\tan\theta \implies \theta = \arctan \left( \frac{x}{z} \right) $$ – Khallil Aug 01 '15 at 23:07
  • Ooooohhhhh, I think I see. So the integral $\int1d\theta=\theta=\arctan\frac{x}{z}$. Thanks for being patient with me! – PoGaMi Aug 01 '15 at 23:13
  • Exactly, @PoGaMi. You got it! (Don't forget the constant of integration) – Khallil Aug 01 '15 at 23:14
  • As a side note, did you know to use that trig sub from experience or are there standard rules? I took Calc 2 about 8yrs ago, so I'm a bit lost in this part of Calc 3. – PoGaMi Aug 01 '15 at 23:15
  • It helped to see an arctan in your answer, but if I ever see something like $bx^2 + c$, I usually try some kind of trig sub to exploit the Pythagorean identities. – Khallil Aug 01 '15 at 23:17
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    @PoGaMi In regards to your last question just above, on how to select the best trig substitution, you might want to take a look at my answer to another question, here. – wltrup Aug 01 '15 at 23:18