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I have been challanged by my teacher to solve this integral, However he gave me no hints, and I have no idea how to start

$$\int_0^1 x\sqrt{\frac{1-x^2}{1+x^2}}dx $$

I noticed that putting $x=1-x$ only expands the term and doesnt really help, So anyone got any other idea, I did try by Parts, It only makes it more messier

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4 Answers 4

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Substitute $x^2\to x$. Then,

$$\int_0^1x\sqrt{\frac{1-x^2}{1+x^2}}\,dx=\frac12 \int_0^1\sqrt{\frac{1-x}{1+x}}\,dx$$

Now, letting $x\to \cos x$ and using $1-\cos x=2\sin^2(x/2)$ and $1+\cos x=2\cos^2(x/2)$ yields

$$\begin{align} \frac12 \int_0^1\sqrt{\frac{1-x}{1+x}}\,dx&=\frac12 \int_0^{\pi/2} \sin x\tan (x/2)\,dx\\\\ &=\int_0^{\pi/2} \sin^2 (x/2)\,dx\\\\ &=\frac12\int_0^{\pi/2}(1-\cos x)\,dx\\\\ &=\pi/4-1/2 \end{align}$$

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    $\begingroup$ The $-1$ in the end should be $-1/2$. $\endgroup$
    – mickep
    Dec 2, 2015 at 17:44
  • $\begingroup$ @mickep Yes it is. Good catch. +1 Thank you!! Edited. -Mark $\endgroup$
    – Mark Viola
    Dec 2, 2015 at 17:47
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HINT:

$$\arccos(x^2)=2y\implies0\le2y\le\pi, x=0\implies2y=\dfrac\pi2, x=1\implies2y=0$$

$$\implies\cos2y=x^2$$ and $$\sqrt{\dfrac{1-x^2}{1+x^2}}=+\tan y$$ as $0\le y\le\dfrac\pi4$

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HINT : $$I=\int_{0}^{1}\frac{x-x^3}{\sqrt{1-x^4}}dx=\int_{0}^{1}\frac{xdx}{\sqrt{1-x^4}}+\frac{1}{4}\int_{0}^{1}\frac{-4x^3dx}{\sqrt{1-x^4}}$$

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  • $\begingroup$ O.o I didn't get this $\endgroup$ Dec 2, 2015 at 17:59
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HINT:

$$\int_{0}^{1}x\sqrt{\frac{1-x^2}{1+x^2}}\space\text{d}x=$$


Substitute $u=x^2$ and $\text{d}u=2x\space\text{d}x$. New lower bound $u=0^2=0$ and upper bound $u=1^2=1$:


$$\frac{1}{2}\int_{0}^{1}\sqrt{\frac{1-u}{1+u}}\space\text{d}u=$$


Substitute $s=\frac{1-u}{1+u}$ and $\text{d}s=\left(-\frac{1-u}{(1+u)^2}-\frac{1}{1+u}\right)\space\text{d}u$.

New lower bound $s=\frac{1-0}{1+0}=1$ and upper bound $s=\frac{1-1}{1+1}=0$:


$$-\int_{1}^{0}\frac{\sqrt{s}}{(-s-1)^2}\space\text{d}s=$$ $$\int_{0}^{1}\frac{\sqrt{s}}{(-s-1)^2}\space\text{d}s=$$


Substitute $p=\sqrt{s}$ and $\text{d}p=\frac{1}{2\sqrt{s}}\space\text{d}s$.

New lower bound $p=\sqrt{0}=0$ and upper bound $p=\sqrt{1}=1$:


$$2\int_{0}^{1}\frac{p^2}{(-p^2-1)^2}\space\text{d}p=$$ $$2\int_{0}^{1}\frac{p^2}{(p^2+1)^2}\space\text{d}p$$

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