Does the next integral $\int_0^{\pi\over 2}{ x \sin(\tan x)\over \cos x}dx$ converge? Does the integral $\displaystyle\int_0^{\pi\over 2}{ x \sin(\tan(x))\over \cos x}dx$ converge? I found it diffcult to solve. Thanks in advance for any assistance! 
 A: The only problem is as
$x \to \pi/2$,
so let's look at
$\int_{{\pi\over 2}-c}^{\pi\over 2}{ x \sin(\tan(x))\over \cos x}dx$
and see what happens for small $c$.
(I use $c$ instead of $\epsilon$ because is takes fewer characters to type.)
Replacing $x$ by $\pi/2-x$,
this becomes
$\int_0^c{ (\pi/2-x) \sin(\tan(\pi/2-x))\over \cos (\pi/2-x)}dx
=\int_0^c{ (\pi/2-x) \sin(1/\tan(x))\over \sin (x)}dx
$.
For small $x$,
$\sin(x) \text{ and }\tan(x) \approx x$
and $\pi/2-x \approx \pi/2$,
so this becomes
$\int_0^c{ (\pi/2) \sin(1/x)\over x}dx
\approx (\pi/2)\int_0^c  \frac{\sin(1/x)}{x} dx
$.
Let $y=1/x$, so $dx=-dy/y^2$.
The integral becomes
$\int_{1/c}^{\infty} \frac{y\sin(y)}{y^2} dy
=\int_{1/c}^{\infty} \frac{\sin(y)}{y} dy
$.
This can be proven to converge in a number of ways.
Here are two:


*

*Divide the region into intervals
where $\sin(x)$ is of constant sign,
show that the absolute values of the integral
over each interval decreases,
and use the alternating series theorem.

*Integrate by parts to get
$ -\cos(y)/y - \int(\cos(y)/y^2)dy$
and since the first part converges
and the second part is dominated by
$\int_{1/c}^{\infty} 1/y^2 dy
= -\frac1{y}\big|_{1/c}^{\infty}
=c
$,
it converges.
