How to show that these integrals converge? What test do I use to show that the following integral converges?
If you could provide me with the process that leads to the answer that would really help.


*

*$\displaystyle  \int_{0}^{1}\frac{x^n}{\sqrt{1-x^4}}\,dx$

*$\displaystyle \int_{0}^{\pi /2}\frac{\ln(\sin(x))}{\sqrt{x}}\,dx$


Thanks
 A: I like to use the comparison tests for these kind of problems.
For the first integral, use the fact that $x^n \leq x$ on $x \in [0,1]$ $\forall n \in \mathbb{N}$, and that $x^4<x^2$ on $[0,1]$, giving 
$$\frac{x^n}{\sqrt{1-x^4}} \leq \frac{x}{\sqrt{1-x^2}}$$
and use the fact that $f\leq g \implies \int f \leq \int g$ on any bounded interval.
Hence,
$$\begin{align}
\int_0^1 \frac{x^n}{\sqrt{1-x^4}} \mathrm{d}x &\leq \int_0^1 \frac{x}{\sqrt{1-x^2}} \mathrm{d}x
\\
&=\left. -\sqrt{1-x^2} +c \right|_0^1
\\
&= 1
\end{align}$$
For the second use a similar idea, except here note that $\ln x < x$ $\forall x>0$ and that $\sin x< x$ for all $x$.
We do all this to get the integral of $g$ on the right side to be one which we can compute in closed form, which gives a bound on the integral of $f$ which you want to show converges, which is of course the standard comparison test. 
A: To check convergence at $x=0$, check that the integrand behaves as $x^a$, where $a>-1$.  If there is something that looks like a singularity away from $x=0$, say at $x=x_0$, then substitute $y=x_0-x$.  Further, $\ln{x}$ represents an integrable singularity.
For the first integral, you need to check the that singularity at $x=1$ is integrable; i.e., that, when you substitute $y=1-x$, then the integral behaves as $y^a$, where $a>-1$ at $y=0$.  That is, check the behavior of
$$[1-(1-y)^4]^{-1/2}$$
at $y=0$.
For the second integral, use the fact that $x^a \ln{x}$ is integrable at $x=0$ when $a>-1$.  How does $\sin{x}$ behave near $x=0$?
