I'm trying to solve or estimate this integral $$ I=\int\limits_{\arcsin{k}}^{\pi/2}\dfrac{1}{(\sin{x})^{1+a}}\mathrm{d}x, $$ where $0<k<1/2$ and $a>0$. The estimate should depend on $k$.
I have tried something like this: since $$ \arcsin{k}\leq x\leq\pi/2\quad\Rightarrow\quad k\leq\sin{x}\leq 1 \quad\Rightarrow\quad 1\leq\dfrac{1}{\sin{x}}\leq\dfrac{1}{k}\quad\Rightarrow\quad 1\leq\dfrac{1}{(\sin{x})^{1+a}}\leq\dfrac{1}{k^{1+a}} $$ we get $$ I\leq\int\limits_{\arcsin{k}}^{\pi/2}\dfrac{1}{k^{1+a}}\mathrm{d}x\leq\dfrac{\pi}{2}k^{-(1+a)}. $$ But this is not good (or "sharp") enough. I need a better estimate or maybe some integration by parts. Any ideas for calculating the integral $I$ ?
Let $$f(x)=\frac{1}{\sin^{1+a} x}-\frac{1}{x^{1+a}},$$ and let $c=\arcsin k$. Then $$\int_c^{\pi/2} \frac{1}{\sin^{1+a} x}dx=\int_c^{\pi/2} \frac{1}{x^{1+a}}\,dx+\int_c^{\pi/2} f(x)\,dx.$$ We can compute the first integral on the right explicitly. It remains to deal with the second.
The function $f(x)$ is reasonably well-behaved, since subtracting $\frac{1}{x^{1+a}}$ has taken the "badness" out of our original function when $k$ is small. So the integral of $f(x)$ can be reliably approximated by one of the usual numerical methods, here applied symbolically. For instance we could use Simpson's Rule with $n=2$.