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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.

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  • $\begingroup$ I have edited the title to reflect the fact that your question seems to be more about calculating the integral, rather than answering a question of convergence. I apologise if I am mistaken; you can always edit the convergence question back in. $\endgroup$ – Will R Dec 28 '16 at 0:09
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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 $(0, \pi/2)$.

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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.

enter image description here

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).

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$$\int_{0}^{\pi/2}\sin\left(\frac{1}{\cos x}\right)\,dx = \int_{0}^{\pi/2}\sin\left(\frac{1}{\sin x}\right)\,dx = \int_{0}^{1}\frac{\sin(1/x)}{\sqrt{1-x^2}}\,dx =\int_{1}^{+\infty}\frac{\sin(x)}{x\sqrt{x^2-1}}$$ 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 $\color{blue}{1}$.

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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 $1$.

Mid-point sum of f

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