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Evaluate $$ \int\frac{\mathrm{d}x}{(1+x^2)\sqrt{1-x^2}}. $$

Set $x=\tan u\implies\mathrm{d}x=\sec^2u\,\mathrm{d}u$. $$ I=\int\frac{\mathrm{d}x}{(1+x^2)\sqrt{1-x^2}}=\int\frac{\sec^2u\,\mathrm{d}u}{\sec^2u\,\dfrac{\sqrt{1-2\sin^2u}}{\cos u}}=\int\frac{\cos u\,\mathrm{d}u}{\sqrt{1-2\sin^2u}}. $$ Set $t=\sqrt2\sin u\implies\mathrm{d}t=\sqrt2\cos u\,\mathrm{d}u$ $$ I=\frac1{\sqrt2}\int\frac{\mathrm{d}t}{\sqrt{1-t^2}}=\frac1{\sqrt2}\sin^{-1}t+c=\frac1{\sqrt2}\sin^{-1}\left(\sqrt{2}\sin u\right)+c=\frac1{\sqrt2}\sin^{-1}\left(\frac{\sqrt2x}{\sqrt{1+x^2}}\right)+c. $$ But my reference gives the solution $\dfrac{-1}{\sqrt{2}}\sin^{-1}\sqrt{\dfrac{1-x^2}{1+x^2}}+c$, where am I going wrong with my attempt ?

Ice Tea
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SOORAJ SOMAN
  • 8,038

3 Answers3

2

Set $x=\sin y,-\dfrac\pi2<x<\dfrac\pi2,\tan y=\dfrac x{\sqrt{1-x^2}}$

$$\int\dfrac{dx}{(1+x^2)\sqrt{1-x^2}}=\int\dfrac{dy}{1+\sin^2y}=\dfrac12\int\dfrac{\sec^2y\ dy}{1/2+\tan^2y}=\dfrac1{\sqrt2}\arctan(\sqrt2\tan y)$$

$$=\dfrac1{\sqrt2}\arctan\dfrac{\sqrt2x}{\sqrt{1-x^2}}$$

If $\arctan\dfrac{\sqrt2x}{\sqrt{1-x^2}}=u,-\dfrac\pi2<u<\dfrac\pi2$

$\tan u=\dfrac{\sqrt2x}{\sqrt{1-x^2}}$

$\sec u=+\sqrt{1+\dfrac{2x^2}{1-x^2}}=\sqrt{\dfrac{1+x^2}{1-x^2}}$

$\sin u=\dfrac{\tan u}{\sec u}=?$

So, your solution is correct

lab bhattacharjee
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To verify $$ I=\int\frac{{d}x}{(1+x^2)\sqrt{1-x^2}}= \frac1{\sqrt2}\sin^{-1}\frac{\sqrt2x}{\sqrt{1+x^2}} $$ utilize an alternative substitution $x=\sinh t$ to arrive at $$I= \int\frac{\text{sech}^2t\ dt}{\sqrt{1-2\tanh^2t}}= \frac1{\sqrt2}\sin^{-1}(\sqrt2\tanh t)=\frac1{\sqrt2}\sin^{-1}\frac{\sqrt2x}{\sqrt{1+x^2}} $$

Quanto
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0

Substitute $x=\sqrt{\dfrac{1-y}{1+y}} \implies y=\dfrac{1-x^2}{1+x^2}$ :

$$\begin{align*} & \int\frac{dx}{(1+x^2)\sqrt{1-x^2}} \\ &= - \int\frac{dx}{\left(1+\frac{1-y}{1+y}\right)\sqrt{1-\frac{1-y}{1+y}}} \, \sqrt{\frac{1+y}{1-y}} \frac{dy}{(1+y)^2} \\ &= -\frac1{2\sqrt2} \int\frac{dy}{\sqrt y\sqrt{1-y}} \\ &= -\frac1{\sqrt2} \int\frac{d\sqrt y}{\sqrt{1-\left(\sqrt y\right)^2}} \\ &= -\frac1{\sqrt2} \arcsin\left(\sqrt y\right) + C \\ &= -\frac1{\sqrt2} \arcsin\left(\sqrt{\frac{1-x^2}{1+x^2}}\right) + C \end{align*}$$

user170231
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