# Why isn't $\int \frac{dx}{(1+x^2)\sqrt{1+x^2}} = - \sin(\arctan(x))$?

Why isn't $\int \frac{dx}{(1+x^2)\sqrt{1+x^2}} = - \sin(\arctan(x))$?

So I understand that if we take the derivative of $-\sin(\arctan(x))$ we won't get what is in the integral. My problem of understanding is within the solution process.

Let $x = \tan(u)$ then $dx = du/\cos^2(u)$.

Since $1 + \tan^2(u) = 1 / \cos^2(u)$, then $dx = (1 + \tan^2(u)) du$.

Thus we have $$\int \frac{1 + \tan^2(u)}{(1 + \tan^2(u))\sqrt{1 + \tan^2(u)}}du$$ or $$\int \frac{du}{\sqrt{1 + \tan^2(u)}}.$$ But $1 + \tan^2(u) = 1/\cos^2(u)$, so $$\int \frac{du}{\sqrt{\frac{1}{\cos^2(u)}}}$$ or $$\int \sqrt{\cos^2(u)}du = \int |\cos(u)|du.$$ And this is where I have a problem. In the case that $\cos(u) \geq 0$ I get that the integral is $\sin(\arctan(x))$. But if $\cos(u) < 0$ then the integral is $-\sin(\arctan(x))$. Which we can tell that is incorrect by taking the derivative. So my understanding is that $\cos(u)$ must be greater or equal to zero, but I can't see why that should be.

• $$\sin\arctan x=\frac x{\sqrt{1+x^2}}\;$$ and thus your solution is correct. – DonAntonio Mar 15 '16 at 22:57

$u = \arctan x$

$-\pi/2 < u < \pi/2$

$\cos u > 0$

And, $\sin(\arctan x) = \dfrac{x}{\sqrt{1+x^2}}$

If we look at the transformation you have carried carefully,

$x = \tan u$. This tranformation is 1 to 1 mapping between x's $(-\infty +\infty)$ and u's $(-\frac{\pi}{2}, \frac{\pi}{2})$. $\cos u > 0$ for $u \in (-\frac{\pi}{2}, \frac{\pi}{2})$. There is no need to consider the case $\cos u < 0$.

This is because cos(u) can't be smaller than (or equal to) 0 when x is a real number.

cos(u) = cos(arctan(x))

You can plot cos(arctan(x)) on a graph if you want to see this.