Finding $\int x \sqrt{1- x^2 \over 1 + x^2}dx$.

Question

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

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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 ?

Answer 1

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 $\arcsin(x^2)$.

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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 $x=1$.

Answer 2

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

$u^2 = 1 + x^2$

$2u du = 2x dx$

$u du = x dx$

Here you are wrong.

Answer 3

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

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

Answer 4

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?