3
$\begingroup$

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.

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

$u = \arctan x$

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

$\cos u > 0$

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

$\endgroup$
2
$\begingroup$

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 $\cos u < 0$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.