Articles of trigonometric integrals

On the integral $\int_0^1\frac{dx}{\sqrtx\ \sqrt{1-x}\ \sqrt{1-x\,\beta^2}}=\frac{2\pi}{7\sqrt{2}\,\beta}$ and $\cos\frac{2\pi}{7}$

(This continues the post on integrals that use roots of reciprocal polynomials.) Given $N=7$. First, how do we show that the algebraic number $\beta$ that solves, $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt[4]{1-x}\ \sqrt{1-x\,\beta^2}}=\frac{1}{7}\,\frac{2\pi}{\sqrt{2}\;|\beta|}\tag1$$ is a root of, $$\beta^2-\frac{2}{1+2\cos\tfrac{\pi}{14}}\beta+\frac{4}{2-\sqrt{2}\,\sec\tfrac{13\pi}{28}}=0$$ Second, the algebraic number $\gamma$ for the related integral, $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{7}\,\frac{2\pi}{\sqrt{2\gamma}}\tag2$$ is a root of, $$\gamma^2-2\cos\tfrac{2\pi}{7}\,\gamma+\frac{2-\sqrt{2}\,\csc\tfrac{9\pi}{28}}{4}=0$$ Equivalently, these roots $\beta$ […]

Is there a way to calculate $\displaystyle \sum_{k=0}^{n}\sin({2^kx})$?

The context for this question is basic curiosity. I know that there are formulas for calculating $\displaystyle \sum_{k=0}^{n}\sin({kx})$ and $\displaystyle \sum_{k=0}^{n}\sin({2kx})$. Is there a formula/method/identity that could be used to find $\displaystyle \sum_{k=0}^{n}\sin({2^kx})$ ?

Area bounded by $\cos x+\cos y=1$

What is the area of the region $\cos x+\cos y > 1$, where $|x|,|y|<\pi$? In other words, is there a “closed” form — using functions that are well-known and nice to work with — for this integral? $$4\int_0^{\pi/2}\cos^{-1}(1-\cos x)\,dx = 7.2948823845\ldots$$ I can approximate it pretty well: >>> scipy.integrate.quad(func=lambda x: math.acos(1-math.cos(x)), … a=0, b=math.pi/2, … […]

Prove $\int_0^1 \frac{4\cos^{-1}x}{\sqrt{2x-x^2}}\,dx=\frac{8}{9\sqrt{\pi}}\left(9\Gamma(3/4)^2{}_4F_3(\cdots)+\Gamma(5/4)^2{}_4F_3(\cdots)\right)$

Mathematica gives the following. But how?! $$\small{\int_0^1 \dfrac{4\cos^{-1}x}{\sqrt{2x-x^2}}\,dx=\frac{8}{9\sqrt{\pi}}\left( 9\Gamma\left(\tfrac{3}{4}\right)^2{}_4F_3\left( \begin{array}{c}\tfrac14\,\tfrac14\,\tfrac34\,\tfrac34\\\tfrac12\,\tfrac54\,\tfrac54\end{array};\tfrac14\right) +\Gamma\left(\tfrac{5}{4}\right)^2{}_4F_3\left( \begin{array}{c}\tfrac34\,\tfrac34\,\tfrac54\,\tfrac54\\\tfrac32\,\tfrac74\,\tfrac74\end{array};\tfrac14\right) \right)}$$ Is there some clever trig substitution that leads to the right-hand side? Or a integral on the complex plane? Can we make it look like a Mahler measure? To aid comparison, we can rewrite this as: $$\small{\int_0^1 \dfrac{4\cos^{-1}x}{\sqrt{2x-x^2}}\,dx =\frac{8\,\Gamma{\left(\frac34\right)}^2}{\sqrt{\pi}}\,{_4F_3}{\left(\frac14,\frac14,\frac34,\frac34;\frac12,\frac54,\frac54;\frac14\right)}\color{red}+\frac{\Gamma{\left(\frac14\right)}^2}{18\sqrt{\pi}}{_4F_3}{\left(\frac34,\frac34,\frac54,\frac54;\frac32,\frac74,\frac74;\frac14\right)}}=\color{red}{\,??}$$ This seems to […]

Calculate $\int\left( \sqrt{\tan x}+\sqrt{\cot x}\right)dx$

This question already has an answer here: Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$? 8 answers

How to evaluate the integral $\int \sqrt{1+\sin(x)} dx$

To find: $$\int \sqrt{1+\sin(x)} dx$$ What I tried: I put $\tan(\frac{x}{2}) = t$, using which I got it to: $$I = 2\int \dfrac{1+t}{(1+t^2)^{\frac{3}{2}}}dt$$ Now I am badly stuck. There seems no way to approach this one. Please give a hint. Also, can we initially to some manipulations on the original integral to make it easy? […]