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Find the indefinite integral

$$\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $$

Is there a smart substitution or algebric trick that I'm missing? Because integration by parts hasn't helped..

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    $\begingroup$ I believe your supposed to substitute a trig function. You might have to guess around on which on though. $\endgroup$ – Zach466920 Oct 30 '15 at 15:36
  • $\begingroup$ I see the derivatives of arctan(x) and arcsin(x) there $\endgroup$ – Stabilo Oct 30 '15 at 15:40
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    $\begingroup$ Hint: substitute $u = \frac{x\sqrt2}{\sqrt{1-x^2}}$ $\endgroup$ – Nicholas Oct 30 '15 at 15:43
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    $\begingroup$ A more natural way to discover the substitution is first set $x = \sin\theta$ and then change variable to $t = \tan\theta = \frac{x}{\sqrt{1-x^2}}$. $\endgroup$ – achille hui Oct 30 '15 at 15:55
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Substitute $u = \frac{x\sqrt2}{\sqrt{1-x^2}}, \frac{dx}{du} =\frac{(1-x^2)^{3/2}}{\sqrt2} $

$\begin{align*} \int {\frac1 {(1+x^2) \sqrt{1-x^2}}dx} &= \frac1{\sqrt2}\int \frac{1-x^2}{1+x^2} du \\ &= \frac1{\sqrt2}\int \frac1 {\frac{1+x^2}{1-x^2}} du \\ &= \frac1{\sqrt2}\int \frac1 {1+\frac{2x^2}{1-x^2}} du \\ &= \frac1{\sqrt2}\int \frac1 {1+ (\frac{x\sqrt2}{\sqrt{1-x^2}})^2} du \\ &= \frac1{\sqrt2}\int \frac1 {1+ u^2} du \\ &= \frac1{\sqrt2} \tan^{-1} \left(\frac{x\sqrt2}{\sqrt{1-x^2}} \right) + C \end{align*}$

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  • $\begingroup$ in du/dx it should be -3/2 ?? and where is the du ? why you leave it dx ? $\endgroup$ – Stabilo Oct 30 '15 at 16:13
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    $\begingroup$ My bad let me edit $\endgroup$ – Nicholas Oct 30 '15 at 16:14
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Let $$I = \int\frac{1}{(1+x^2)\sqrt{1-x^2}}dx\;,$$ Now Put $\displaystyle x= \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$

So we get $$I = \int\frac{t^3}{(1+t^2)\sqrt{t^2-1}}\cdot -\frac{1}{t^2}dt = -\int\frac{t}{(1+t^2)\sqrt{t^2-1}}dt$$

Now Put $(t^2-1)=u^2\;,$ Then $2tdt = 2udu\Rightarrow tdt = udu$

So we get $$I = -\int\frac{u}{u^2+2}\cdot \frac{1}{u}du = -\int\frac{1}{u^2+2}du$$

So we get $$I = -\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)+\mathcal{C}=-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{t^2-1}}{\sqrt{2}}\right)+\mathcal{C}$$

So we get $$I = \int\frac{1}{(1+x^2)\sqrt{1-x^2}}dx=-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{1-x^2}}{\sqrt{2}x}\right)+\mathcal{C}$$

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  • $\begingroup$ great answer! but he answered first =\ ... $\endgroup$ – Stabilo Oct 30 '15 at 16:31
  • $\begingroup$ Stabilo I also like Nicholas Answer. $\endgroup$ – juantheron Oct 30 '15 at 16:45
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Hint:

$$\int {dx \over {(1+x^2) \sqrt{1-x^2}}}=\int\frac{d\arcsin(x)}{1+\sin^2(\arcsin(x))}=\int\frac{dt}{1+\sin^2(t)}=\int\frac {dz}{iz\left(1+\left(\dfrac{z-z^{-1}}{2i}\right)^2\right)},$$ ($t=\arcsin(x),z=e^{it}$) and you have a rational expression.

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  • $\begingroup$ square root got lost? $\endgroup$ – GEdgar Oct 30 '15 at 16:41
  • $\begingroup$ No, was extraneous, fixed. Thanks. $\endgroup$ – Yves Daoust Oct 30 '15 at 16:52
  • $\begingroup$ @YvesDaoust .Sharp substitution. But that integral with the $1+sin^2t$ in the denom can of course be done without the complex stuffff. $\endgroup$ – imranfat Nov 1 '15 at 1:10
  • $\begingroup$ @imranfat: the goal is to let the transcendental functions completely disappear to obtain a rational expression, for which a systematic procedure is known. $\endgroup$ – Yves Daoust Nov 1 '15 at 10:53
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Let $x=\sin\theta\implies dx=\cos\theta \ d\theta$, $$\int \frac{dx}{(1+x^2)\sqrt{1-x^2}}$$ $$=\int \frac{\cos\theta\ d\theta}{(1+\sin^2\theta)\cos\theta}$$ $$=\int \frac{\sec^2\theta\ d\theta}{\sec^2\theta(1+\sin^2\theta)}$$ $$=\int \frac{\sec^2\theta\ d\theta}{\sec^2\theta+\tan^2\theta}$$ $$=\int \frac{\sec^2\theta\ d\theta}{1+2\tan^2\theta}$$ $$=\int \frac{ d(\tan\theta)}{1+(\sqrt2 \tan\theta)^2}$$ $$=\frac{1}{\sqrt 2}\tan^{-1}\left(\sqrt 2\tan\theta\right)+C$$ $$=\color{red}{\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{x\sqrt 2}{\sqrt{1-x^2}}\right)+C}$$

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