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I am trying to find $$\int \dfrac{\sqrt{1-x^2}}{x^2} \operatorname d \! x$$

I can get to $${\int \dfrac{\sin x}{\cos^2x}\operatorname d \! x}$$

I tried u substitution by making $u=\cos x$ and $du= −\sin x dx$

but this leads to $$\int \dfrac{1}{u^2}\operatorname d \! u$$ and this is where I get stuck $$ \dfrac{1}{\cos x}.$$

The actual answer is $\int \frac{\sqrt{1-x^2}}{x^2}dx=-\frac{\sqrt{1-x^2}}{x}-\arcsin \left(x\right)+C$

From where does the arcsin(x) come from? Also the first term doesn't make sense if I'm dealing with 1/cosx ... :(

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Careful, you left out your variable of integration (the $dx$) and also used $x$ to mean two different things. Let's back up and be more careful. $$x = \cos \theta,\ dx = -\sin \theta d\theta$$ $$\int \frac{\sqrt{1-x^2}}{x^2} dx = \int \frac{\sin \theta}{\cos^2 \theta}(-\sin \theta d\theta) = -\int\tan^2\theta d\theta$$ (Since I wrote this answer other people have edited the original question and added in the $d$s, but the other problems remain.)

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  • $\begingroup$ you are write i will edit my answer $\endgroup$ – Semsem Feb 28 '14 at 21:50
  • $\begingroup$ Yes this is what I'm looking for. I had the right variables on paper but I forgot about the dx, thanks for pointing that out. Now I just have to figure out what the integral of tan^2 is haha $\endgroup$ – zman Feb 28 '14 at 22:00
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    $\begingroup$ There's a trig identity involving $\tan^2\theta$ that will probably be useful. $\endgroup$ – dfan Feb 28 '14 at 22:03
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You're almost there. $$\int \frac{du}{u^2}=\int u^{-2}du=-u^{-1}+C$$

Now substitute back to your original variables.

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  • $\begingroup$ That's exactly where I'm stuck $$ \dfrac{1}{cos x}$$ what can I substitute back in there to get the answer $\endgroup$ – zman Feb 28 '14 at 21:43
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Let $$x = \cos \theta $$ $$dx = - \sin \theta \ d \theta $$

then you should get $$\int \frac{ \sin \theta}{\cos^2 \theta} (- \sin \theta) \ d \theta $$ $$ \int (1 - \sec^2 \theta) \ d\theta $$

You should be able to finish it now.

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