Calculating $\int_{0}^{\frac{\pi}{2}}\sin(\sec x)\,dx$ I want to calculate
$$\int_{0}^{\frac{\pi}{2}}\sin(\sec x)\,dx.$$
I couldn't really figure out. Should I do integration by parts? I can't calculate the integral this way.
 A: You can't find the antiderivative of this function in terms of elementary functions, but you do know it converges because it is bounded and continuous on $(0, \pi/2)$. 
A: There does not seem to be a closed form expression for this integral, but here is a way to show it converges (note that this is indeed an improper integral, as secant is undefined at $\pi/2$).
Here is a picture of the graph from wolfram alpha. Note that it behaves similarly to $\sin(\tfrac{1}{x})$ for similar reasons.

Now to see why the function converges, note that $-1\leq \sin(\sec(x))\leq 1$, so 
$$
0\leq \sin(\sec(x))+2\leq 2.
$$
In particular, 
$$
\int_0^{a}(\sin(\sec(x))+1)dx\leq 2a
$$
for all $a<\pi/2$. 
Further, if $a_1<a_2$, then 
$$
\int_0^{a_1}(\sin(\sec(x))+1)dx\leq \int_0^{a_2}(\sin(\sec(x))+1)dx
$$
as the integrand is nonnegative. 
Thus $\int_0^{a}(\sin(\sec(x))+1)dx$ is an increasing, bounded function of $a$ for $a<\pi/2$, so 
$$
\int_0^{\pi/2}(\sin(\sec(x))+1)dx=\lim_{a\to(\pi/2)^{-}}\int_0^{a}(\sin(\sec(x))+1)dx
$$
exists, and hence the original limit will converge as well (by subtracting where appropriate).
A: $$\int_{0}^{\pi/2}\sin\left(\frac{1}{\cos x}\right)\,dx = \int_{0}^{\pi/2}\sin\left(\frac{1}{\sin x}\right)\,dx = \int_{0}^{1}\frac{\sin(1/x)}{\sqrt{1-x^2}}\,dx =\int_{1}^{+\infty}\frac{\sin(x)}{x\sqrt{x^2-1}}$$
and the last integral is convergent by Dirichlet's test: $\sin x$ is a function with a bounded primitive and $\frac{1}{x\sqrt{x^2-1}}$ is a function decreasing to zero on $(1,+\infty)$. $\frac{\sin(x)}{x\sqrt{x^2-1}}$ is also integrable in absolute value, since $\left|\sin x\right|\leq 1$ and $\frac{1}{x\sqrt{x^2-1}}\approx\frac{1}{x^2}$. 
Numerically, the value of the integral is pretty close to $\color{blue}{1}$.
A: You can compute it numerically, but you cannot use the right end-point because $\sin\left(\sec\frac\pi2\right)$ is not defined. Use the mid-point rule. Let $n$ be the number of partitions and $\Delta x$ be the length of each interval. Then, $n\Delta x=\frac\pi2$. Let $f(x)=\sin\left(\sec x\right)$. The mid-point sum is
$$M=\Delta x\left(f\left(\frac{\Delta x}{2}\right)+f\left(\frac{3\Delta x}{2}\right)+f\left(\frac{5\Delta x}{2}\right)+\dots+f\left(\frac{\pi-\Delta x}{2}\right)\right)$$
Substituting $\Delta x=\frac{\pi}{2n}$, we have
$$M=\frac{\pi}{2n}\left(f\left(\frac{\pi}{4n}\right)+f\left(\frac{3\pi}{4n}\right)+f\left(\frac{5\pi}{4n}\right)+\dots+f\left(\frac{(2n-1)\pi}{4n}\right)\right)$$
The result is very near $1$.

