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How should I integrate $\int _{ 0 }^{ \frac{1}{\sqrt{3}} } \frac { dx }{ (1+x^2)\sqrt{1-x^2}\ } $ ?

I tried substituting $x=1/t$,but it is not working.Could someone help?Thanks.

  • Try $x=\cos t$. – vadim123 Nov 17 '15 at 15:04
  • Does this help: Try $x = \sqrt{\cos(2\theta)}$ then you have $1-x^{2}= 1-\cos(2\theta)=2 \sin^{2}\theta$ and $1+x^{2} = 2\cos^{2}\theta$?? – C.S. Nov 17 '15 at 15:10
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    For other alternatives, see http://math.stackexchange.com/questions/1505148/find-the-indefinite-integral-int-dx-over-1x2-sqrt1-x2/1505197#1505197 – Nicholas Nov 17 '15 at 15:15

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Let $x=\sin \theta$ and then \begin{eqnarray} \int _{ 0 }^{ \frac{1}{\sqrt{3}} } \frac { dx }{ (1+x^2)\sqrt{1-x^2}\ }&=&\int _{ 0 }^{ \arcsin\frac{1}{\sqrt{3}} } \frac {d\theta }{1+\sin^2\theta}\\ &=&\frac{1}{\sqrt 2}\arctan(\sqrt{2}\tan\theta)\bigg|_{ 0 }^{\arcsin\frac{1}{\sqrt{3}}}\\ &=&\frac{1}{\sqrt 2}\arctan(\sqrt{2}\cdot\frac{1}{\sqrt2})\\ &=&\frac{\pi}{4\sqrt 2}. \end{eqnarray} Here $\arctan(\arcsin\frac{1}{\sqrt{3}})=\frac{1}{\sqrt2}$ and $$ \int\frac{1}{1+\sin^2\theta}d\theta=\int\frac{1}{1+2\tan^2\theta}d\tan\theta=\frac{1}{\sqrt2}\arctan(\sqrt2\tan\theta)+c.$$

xpaul
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