Checking the convergence of $\int _{-1}^{1} \frac{2^{\arcsin x}}{1-x}\, dx$ and an another integral I need to check convergence of 


*

*$$\int _{-1}^{1} \frac{2^{\arcsin x}}{1-x}\, dx$$

*$$\int _0^{\frac{\pi }{2}}\:\frac{x}{\cos x}\cdot \sin\left(\tan\left(x\right)\right)dx$$
My attempt:
For the first one, what I did :
$$0\le \frac{2^{\arcsin x}}{1-x}\le \frac{2^2}{\left(1-x\right)^1}$$
so the integral convergences.
Is this correct, and any idea how to check the second one?
Thanks a lot.
 A: $\bullet$ For the first one, the inequality you used does not justify the convergence around $x=1^{-}$. However, to study around $1$ you can change to $x=1-h$. Then
$$
\frac{2^{\text{arcsin}\left(x\right)}}{1-x}\underset{h \rightarrow 0}{\sim}\frac{2^{\, \pi/2}}{h}
$$
However $\displaystyle h \mapsto \frac{1}{h}$ diverges around $0$ so as your integral.
$\bullet$ To help you with the second one, you can analyze instead the following integral
$$
\int_{0}^{+\infty}\frac{\sin\left(v\right) \text{arctan}\left(v\right)}{\left(1+v^2\right)\cos\left(\text{arctan}\left(v\right)\right)}\text{d}v=\int_{0}^{+\infty}\frac{\sin\left(v\right) \text{arctan}\left(v\right)}{\sqrt{1+v^2}}\text{d}v
$$
The function within the integral is continous in $\left[0,+\infty\right[$ and satisfies
$$
\left|\frac{\sin\left(v\right) \text{arctan}\left(v\right)}{\sqrt{1+v^2}}\right|\underset{(+\infty)}{\sim}\frac{\pi}{2}\frac{\sin\left(v\right)}{v}
$$
Then, it is semi-convergent as Dirichlet's integral.
By changing variable $x=\text{arctan}\left(v\right)$, you found it is equal to your second integral.
