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

Question

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..

Answer 1

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

Answer 2

Let $x=\sin\theta\implies dx=\cos\theta \ d\theta$, $$\int \frac{dx}{(1+x^2)\sqrt{1-x^2}}$$ $$=\int \frac{\cos\theta\ d\theta}{(1+\sin^2\theta)\cos\theta}$$ $$=\int \frac{\sec^2\theta\ d\theta}{\sec^2\theta(1+\sin^2\theta)}$$ $$=\int \frac{\sec^2\theta\ d\theta}{\sec^2\theta+\tan^2\theta}$$ $$=\int \frac{\sec^2\theta\ d\theta}{1+2\tan^2\theta}$$ $$=\int \frac{ d(\tan\theta)}{1+(\sqrt2 \tan\theta)^2}$$ $$=\frac{1}{\sqrt 2}\tan^{-1}\left(\sqrt 2\tan\theta\right)+C$$ $$=\color{red}{\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{x\sqrt 2}{\sqrt{1-x^2}}\right)+C}$$

Answer 3

Substitute $u = \frac{x\sqrt2}{\sqrt{1-x^2}}, \frac{dx}{du} =\frac{(1-x^2)^{3/2}}{\sqrt2} $

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

Answer 4

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.

Answer 5

Substitute instead $y=\frac x{\sqrt{1+x^2}}$ to integrate $$\int {dx \over {(1+x^2) \sqrt{1-x^2}}} = \int \frac{dy}{\sqrt{1-2y^2}}=\frac1{\sqrt2}\sin^{-1}\frac{\sqrt2 x}{\sqrt{1+x^2}}+C $$

Answer 6

Since the derivative of $\tan^{-1}x$ is present in the integrand, substitute $x=\tan\theta, \theta\in\left(\frac{-\pi}2,\frac\pi2\right)$:

$$\begin{align}\int\frac{\mathrm dx}{(1+x^2)\sqrt{x^2-1}}&=\int\frac{\mathrm d\theta}{\sqrt{1-\tan^2\theta}}\\&=\int\frac{\cos\theta\mathrm d\theta}{\sqrt{1-2\sin^2\theta}}\\&=\frac1{\sqrt2}\sin^{-1}(\sqrt2\sin\theta)+C\\&=\frac1{\sqrt2}\sin^{-1}\left(\frac{\sqrt2x}{\sqrt{1+x^2}}\right)+C\end{align}$$