Check solution of problem $\lim_{n\to\infty}\int_0^\frac{\pi}{2} \frac{\sin^nx}{\sqrt{1+x}}\, dx = 0$ I need check my solution of problem: 
$$\lim_{n\to\infty}\int_0^\frac{\pi}{2} \frac{\sin^nx}{\sqrt{1+x}}\, dx = 0$$

Solution: 
$$0 \leq\underset{n\to \infty }{\text{lim}}\int_0^{\frac{\pi }{2}} \frac{\sin ^n(x)}{\sqrt{x+1}} \, dx  
\leq  \underset{n\to \infty }{\text{lim}}\int_0^{\frac{\pi}{2}} \sin^n(x) dx = \underset{n\to \infty }{\text{lim}}\int_0^{\frac{\pi}{2}} \cos^n(x) dx
$$
I apply 
$$ \cos(x) = \frac{e^{i x}+e^{-i x}}{2}$$
$$
\underset{n\to \infty }{\text{lim}}\int_0^{\frac{\pi}{2}} \cos^n(x) dx 
\leq \int_{0}^{\frac{\pi}{2}} e^{n i x} dx $$
because 
$$ e^{-x i} <  e^{x i} $$ for $x \in (0, \frac{\pi}{2})$
so 
$$ 
\int_{0}^{\frac{\pi}{2}} e^{n i x} dx = \frac{1}{in}\exp(i x n)\Bigm|_0^{\frac{\pi}{2}} = 0
$$
but it's greater than 0, which means that the result is 0 .
 A: There is a big issue with your solution!
There is no order relation in the complex field $\mathbb C$. Hence
$$e^{-x i} <  e^{x i}$$ doesn't make sense!
If you know Lebesgue integral, you can use Dominated Convergence theorem which leads to the result in a straight way as $\sin^n (x)$ converges pointwise to $0$ on $[0,\pi/2)$ and the map you consider is bounded by one.
If you only know Riemann integral, it is a bit more technical. Just split the integral
$$\int_0^{\pi/2} = \int_0^{\pi/2 - \epsilon} +\int_{\pi/2 - \epsilon}^{\pi/2}$$
and use the standard definition of a limit.
A: Your proof does not make sense.  For a valid proof use the fact that $\sin $ is increasing for $0<x<\pi /2$. So $\int_0^{\pi/2 -r} \frac {sin^{n} x} {\sqrt {1+x}}dx\leq \int_0^{\pi/2-r} s^{n}dx=s^{n}(\pi /2 -r) \to 0$ for any $r >0$, where $s=\sin (\pi/2-r)$.  [Note that $s <1$]. Now $\int_{\pi/2-r} ^{\pi /2} \frac {sin^{n} x} {\sqrt {1+x}}dx\leq r$ since the integrand is bounded by $1$. Can you finish the proof now?
A one line proof can be given using DCT but I have provided a proof that does not use measure theory. 
A: \begin{align}
\int_0^{\pi/2}\sin^nx\,dx&=\frac12 \text{Beta}\left(\frac{n+1}2\frac12\right) \\
&=\frac12 \frac{\Gamma\left(\frac{n+1}2\right)\Gamma\left(\frac12\right)}{\Gamma\left(1+\frac n2\right)}\\
&=\frac{\sqrt{\pi}}{\sqrt2}n^{-1/2}+O\left(n^{-3/2}\right)
\end{align}
