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$$J =\int x \sqrt{1- x^2 \over 1 + x^2}dx$$


Substituting $u = \sqrt{1 + x^2}$

$$J = \int \sqrt{2 - u^2} du = \sqrt{2}\int \sqrt{1 - \left({u \over \sqrt{2}}\right)^2} du$$

Now substituting $\sin z = u/\sqrt{2}$

$$J = 2\int \cos^2 z dz = z + (\sin 2z)/2 + C = \arcsin\left(\sqrt{1+ x^2}\over 2\right) + {\sqrt{1-x^4}\over 2} + C $$.

The give answer is $\displaystyle \arcsin(x^2) + {\sqrt{1-x^4}\over 2} + C$.

What went wrong in my attempt ?

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  • $\begingroup$ A---B My edit was to simply add a non-mathjax component to the title. No big deal! :-) $\endgroup$ – Namaste Apr 17 '17 at 17:38
  • $\begingroup$ @amWhy Thank you amWhy. $\endgroup$ – A---B Apr 17 '17 at 17:40
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Note that $$\frac{d}{dx}\arcsin(x^2)=\frac{2x}{\sqrt{1-x^4}}$$ and $$\frac{d}{dx}2\arcsin\left(\frac{\sqrt{1+x^2}}{\sqrt 2}\right)=\frac{2x}{\sqrt{1-x^4}}$$

So, the error in the OP is in the term $\arcsin\left(\frac{\sqrt{1+x^2}}{2}\right)$, which should be instead $\arcsin\left(\frac{\sqrt{1+x^2}}{\sqrt 2}\right)$ and the "given answer" needs to have $\frac12 \arcsin(x^2)$ instead of $\arcsin(x^2)$.


The resolution to the apparent discrepancy comes from the identity

$$2\arcsin\left(\sqrt{1+x^2}\over\sqrt{2}\right) = \arcsin(x^2)+\pi/2\tag 1$$

which can be verified directly by taking the sine of both sides of $(1)$ and using $\arcsin(1)=\pi/2$. We have used instead that the two sides of $(1)$ have equal derivatives and are equal at $x=1$.

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  • $\begingroup$ But it still does not match the given answer. $\endgroup$ – A---B Apr 17 '17 at 17:48
  • $\begingroup$ The given answer needs a $1/2$ in front of $\arcsin(x^2)$. $\endgroup$ – Mark Viola Apr 17 '17 at 17:51
  • $\begingroup$ Honestly I don't know how to transform $$\arcsin\left(\sqrt{1+x^2}\over\sqrt{2}\right) = (1/2)\arcsin(x^2)$$. $\endgroup$ – A---B Apr 17 '17 at 17:55
  • $\begingroup$ Well I can buy it since I don't know inverse trig identities.Thank you. $\endgroup$ – A---B Apr 17 '17 at 18:01
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    $\begingroup$ Note that $$\arcsin\left(\sqrt{1+x^2}\over\sqrt{2}\right) = (1/2)\arcsin(x^2)+\pi/4$$ $\endgroup$ – Mark Viola Apr 17 '17 at 18:18
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$u = \sqrt {1+x^2}$

$u^2 = 1 + x^2$

$2u du = 2x dx$

$u du = x dx$

Here you are wrong.

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  • $\begingroup$ What is wrong with that transformation? $\endgroup$ – Mark Viola Apr 17 '17 at 17:33
  • $\begingroup$ I got $\displaystyle u^\prime = 1/(2\sqrt{1+ x^2}) \times 2x = x/\sqrt{1+x^2}$. $\endgroup$ – A---B Apr 17 '17 at 17:34
  • $\begingroup$ @Salahamam_Fatima Why I am pretty sure it is $cos^2 x$. $$\sqrt{2} \int \sqrt{1 - \sin^2 z}\ dz \cos z \sqrt{2} = 2\cos^2 z dz$$ $\endgroup$ – A---B Apr 17 '17 at 17:45
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Put $t=x^2$

and

$$\frac {1-t}{1+t}=u^2$$

or

$$t=\frac{1-u^2}{1+u^2} $$

$$=\frac {2}{1+u^2}-1$$

and

$$dt=\frac{-4u}{(1+u^2)^2} $$

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    $\begingroup$ Yes thank you but I would like to know what did I do wrong ? $\endgroup$ – A---B Apr 17 '17 at 17:36
  • $\begingroup$ Thank you for the alternative method. $\endgroup$ – A---B Apr 17 '17 at 18:05
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HINT:

Let $x^2=\cos2y$ as $x^2\ge0,0\le2y\le\dfrac\pi2\ \ \ \ (1)$

$\implies x\ dx=-\sin2y\ dy,\sin2y=+\sqrt{1-x^4}$

Now $\sin2y=2\sin y\cos y$

and $\sqrt{\dfrac{1-x^2}{1+x^2}}=+\tan y\text{ by } (1)$

Can you take it from here?

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  • $\begingroup$ Yes thank you for this method. I tried this but was quite long to do. $\endgroup$ – A---B Apr 17 '17 at 18:12
  • $\begingroup$ @A---B, $$\int(-2\sin^2y)dy=\int(\cos2y-1)dy=\dfrac{\sin2y}2-y+K$$ Replace the values of $\sin2y, y$ $\endgroup$ – lab bhattacharjee Apr 17 '17 at 18:14
  • $\begingroup$ Oh you are correct I messed it by substituting $t = \cos 2u$ in a intermediate step. Integrals are difficult :(. $\endgroup$ – A---B Apr 17 '17 at 18:20

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