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Using the substitution $x=a\sin\theta$, or otherwise, find $\int\frac{1}{x^2\sqrt{a^2-x^2}}dx$.

My attempt, $x=a\sin\theta$

$dx=a\cos (\theta)d\theta$. Then $\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin ^2(\theta)}$ The given answer is $-\frac{\sqrt{a^2-x^2}}{a^2x}+c$

How to proceed then?

tired
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Mathxx
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3 Answers3

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Other possibility:

Use $x=1/y$, the integral becomes: $$ I(a)=\int -\frac{1}{y^2}\frac{1}{\frac{1}{y^2} \sqrt{a^2-\frac{1}{y^2}}}dy=-\int \frac{y}{\sqrt{a^2y^2-1}}dy=-\frac{1}{a^2}\int \partial_y\left(\sqrt{a^2y^2-1}\right)dy=-\frac{\sqrt{a^2y^2-1}}{a^2}+C $$

Resubstitute: $$ I(a)=-\frac{\sqrt{a^2-x^2}}{ a^2 x}+C $$

tired
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Using your substition:

$$\int\frac1{x^2\sqrt{a^2-x^2}}\;\rightarrow\;\int\frac{a\cos\theta\;d\theta}{a^2\sin^2\theta\cdot a\cos\theta}=-\frac1{a^2}\int\frac{d\theta}{-\sin^2\theta}=-\frac1{a^2}\cot\theta+C$$

Timbuc
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  • How $\sqrt{a^2-a^2\sin ^2(\theta)}$ becomes $a\cos \theta$? – Mathxx Apr 21 '15 at 08:42
  • @Mathxx But... you wrote that in your question before you edited it! Anyway, it's simple trigonometry: $$\sqrt{a^2-a^2\sin^2\theta}=\sqrt{a^2(1-\sin^2\theta)}=a\sqrt{\cos^2\theta}=a \cos\theta$$ – Timbuc Apr 21 '15 at 08:51
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    To be fully correct (since we do not know about intervals or sign of $a$) we should add an absolute value around the $a\cos\theta$, so $\sqrt{a^2-a^2\sin^2\theta}=|a\cos\theta|$. – mickep Apr 21 '15 at 09:03
  • @mickep You've got a point. +1 – Timbuc Apr 21 '15 at 09:05
  • How to proceed then? – Mathxx Apr 21 '15 at 09:05
  • @Mathxx Well, the sign of $;a;$ is unimportant, but the one of $;\cos \theta;$ is. When you ge to Riiemann Integra (i.e., definite integrals), you will have to separate the integration interval in parts where cosine is poisitve and negative. In this case, I think that pointing out what mickep mentioned is enough. – Timbuc Apr 21 '15 at 09:07
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In fact, under $t=\frac{\sqrt{a^2-x^2}}{x}$ , $$\int\frac{dx}{x^2\sqrt{a^2-x^2}}=\int\frac{1}{\frac{a^3t}{(t^2+1)^{3/2}}}\left(-\frac{at}{(t^2+1)^{3/2}}dt\right)\\=-\frac{1}{a^2}\int dt=-\frac{1}{a^2}t+C=-\frac{\sqrt{a^2-x^2}}{a^2x}$$

Antony Theo.
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