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integrate $$\int \frac {1}{(x^2+R^2)^{3/2}}dx $$

This came up doing a physics task, but I have no idea how to integrate it without straight using integration table. I tried to do it integrating by parts but that get me nowhere.

Lugi
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1 Answers1

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A start: Let $x=R\tan\theta$. After a while, you will be integrating $\cos\theta$.

More: Then $dx=R\sec^2\theta \,d\theta$. Note that $$R^2+x^2=R^2+R^2\tan^2\theta=R^2(1+\tan^2\theta)=R^2\sec^2\theta.$$ It follows that $(R^2+x^2)^{3/2}=R^3\sec^3\theta$. Thus $$\int \frac{1}{(R^2+x^2)^{3/2}}\,dx=\int \frac{R\sec^2\theta}{R^3\sec^3\theta}\,d\theta=\int \frac{1}{R^2}\cos \theta\,d\theta.$$

André Nicolas
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  • Just what we physics students are taught as a rule for this repeatedly occurring integral, so much that I almost remembered the results for various limits sometime back. – RE60K Sep 02 '14 at 14:13
  • Because of inverse distance laws, this sort of thing does come up often. For me, it feels better to deal with an integral that is "real" than with something artificial that succombs to a "clever" trick. – André Nicolas Sep 02 '14 at 14:27
  • Im stuck at $\int \frac {1-\tan \theta}{\sqrt{1+\tan\theta }} d\theta $ . What do I do from here on? – Lugi Sep 02 '14 at 19:08
  • I will type a few additional lines in the answer, typing formulas in a comment often gives me TeX problems. – André Nicolas Sep 02 '14 at 20:11
  • Ok, this integral equals to $\frac {\sin\theta}{R^2}$ but what should I do with it to get x involved? The best I can come up with is $\frac {x \cos\theta}{R^3}$ – Lugi Sep 02 '14 at 20:50
  • Draw a right-angled triangle. Label one of the "legs" $x$, and another leg $R$. Label the angle opposite side $x$ using the label $\theta$. Then $x=R\tan\theta$. The hypotenus is $\sqrt{R^2+x^2}$, so $\sin\theta=\frac{x}{\sqrt{R^2+x^2}}$. – André Nicolas Sep 02 '14 at 20:56