¿Cómo calcular los dos integrales?

Question

Cómo calcular las siguientes integrales $$I=\int_0^{\pi/2}x^2\frac{(1-\tan\,x)\sin4x}{\sqrt{\tan x}}dx$ $ $$K=\int_0^{\pi/2}x^2\frac{(1+\tan\,x)\sin4x}{\sqrt{\tan x}}dx$ $

EDITAR

Parece que una manera obvia de atacar este integral es introducir un parámetro $\alpha$ en el argumento del seno. Haciendo esto, sin embargo, hará que la integral a ser indefinido. Tiene que haber un método diferente de ataque.

Answer 1

Estoy empezando a pensar que la primera integral puede ser factible después de todo, al menos en términos de funciones hipergeométricas. He aquí lo que tengo hasta ahora.

$$\begin{align} \mathcal{I} &:=\int_{0}^{\frac{\pi}{2}}x^2\frac{\left(1-\tan{x}\right)\sin{4x}}{\sqrt{\tan{x}}}\,\mathrm{d}x\\ &\approx 0.998987.\\ \end{align}$$

$$\begin{align} \mathcal{I} &=\int_{0}^{\frac{\pi}{2}}x^2\frac{\left(1-\tan{x}\right)\sin{4x}}{\sqrt{\tan{x}}}\,\mathrm{d}x\\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(y+\frac{\pi}{4}\right)^2\frac{\left(1-\tan{\left(y+\frac{\pi}{4}\right)}\right)}{\sqrt{\tan{\left(y+\frac{\pi}{4}\right)}}}\sin{\left[4\left(y+\frac{\pi}{4}\right)\right]}\,\mathrm{d}y;~~[y=x-\frac{\pi}{4}]\\ &=-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(y+\frac{\pi}{4}\right)^2\frac{\left(1-\tan{\left(y+\frac{\pi}{4}\right)}\right)}{\sqrt{\tan{\left(y+\frac{\pi}{4}\right)}}}\sin{\left(4y\right)}\,\mathrm{d}y\\ &=-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(y+\frac{\pi}{4}\right)^2\frac{-\sqrt{2}\sin{\left(y\right)}\csc{\left(\frac{\pi}{4}-y\right)}}{\sqrt{\tan{\left(y+\frac{\pi}{4}\right)}}}\sin{\left(4y\right)}\,\mathrm{d}y\\ &=+\sqrt{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(y+\frac{\pi}{4}\right)^2\frac{\sin{\left(y\right)}\csc{\left(\frac{\pi}{4}-y\right)}}{\sqrt{\tan{\left(y+\frac{\pi}{4}\right)}}}\sin{\left(4y\right)}\,\mathrm{d}y\\ &=\sqrt{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(y+\frac{\pi}{4}\right)^2\frac{\sec{\left(\frac{\pi}{4}+y\right)}}{\sqrt{\tan{\left(y+\frac{\pi}{4}\right)}}}\sin{\left(y\right)}\sin{\left(4y\right)}\,\mathrm{d}y\\ &=\sqrt{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(y+\frac{\pi}{4}\right)^2\sqrt{2\sec{\left(2y\right)}}\sin{\left(y\right)}\sin{\left(4y\right)}\,\mathrm{d}y\\ &=2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(y+\frac{\pi}{4}\right)^2\sqrt{\sec{\left(2y\right)}}\sin{\left(y\right)}\sin{\left(4y\right)}\,\mathrm{d}y\\ &=2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(y^2+\frac{\pi}{2}\,y+\frac{\pi^2}{16}\right)\sqrt{\sec{\left(2y\right)}}\sin{\left(y\right)}\sin{\left(4y\right)}\,\mathrm{d}y\\ &=2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(y^2+\frac{\pi^2}{16}\right)\sqrt{\sec{\left(2y\right)}}\sin{\left(y\right)}\sin{\left(4y\right)}\,\mathrm{d}y\\ &=4\int_{0}^{\frac{\pi}{4}}\left(y^2+\frac{\pi^2}{16}\right)\sqrt{\sec{\left(2y\right)}}\sin{\left(y\right)}\sin{\left(4y\right)}\,\mathrm{d}y\\ &=\frac12\int_{0}^{\frac{\pi}{2}}\left(z^2+\frac{\pi^2}{4}\right)\sqrt{\sec{\left(z\right)}}\sin{\left(\frac{z}{2}\right)}\sin{\left(2z\right)}\,\mathrm{d}z;~~[z=2y]\\ &=\frac12\int_{0}^{\frac{\pi}{2}}\left(z^2+\frac{\pi^2}{4}\right)\sqrt{\frac{\sec{\left(z\right)}-1}{2}}\cdot2\sin{\left(z\right)}\cos{\left(z\right)}\,\mathrm{d}z\\ &=\frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\left(z^2+\frac{\pi^2}{4}\right)\cos{\left(z\right)}\sqrt{\sec{\left(z\right)}-1}\cdot\sin{\left(z\right)}\,\mathrm{d}z\\ &=\frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\left(z^2+\frac{\pi^2}{4}\right)\sqrt{\cos{\left(z\right)}-\cos^2{\left(z\right)}}\cdot\sin{\left(z\right)}\,\mathrm{d}z\\ &=\frac{\pi^2}{4\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\sqrt{\cos{\left(z\right)}-\cos^2{\left(z\right)}}\cdot\sin{\left(z\right)}\,\mathrm{d}z\\ &~~~~~+\frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}}z^2\sqrt{\cos{\left(z\right)}-\cos^2{\left(z\right)}}\cdot\sin{\left(z\right)}\,\mathrm{d}z\\ &=:\frac{\pi^2}{4\sqrt{2}}\,I_{1}+\frac{1}{\sqrt{2}}\,I_{2}.\\ \end{align}$$

El trigonométricas factor de incidencia en el integrands de $I_1$$I_2$, en realidad tiene una escuela primaria antiderivada, que vamos a denominar como $g{(z)}$, que se puede encontrar muy directamente. En particular, esto permite a $I_1$ a evaluarse directamente (alternativamente $I_1$ puede ser evaluado como una función beta).

$$\begin{align} I_{1} &=\int_{0}^{\frac{\pi}{2}}\sqrt{\cos{\left(z\right)}-\cos^2{\left(z\right)}}\cdot\sin{\left(z\right)}\,\mathrm{d}z\\ &=-\int_{1}^{0}\sqrt{t-t^2}\,\mathrm{d}t;~~[t=\cos{z}]\\ &=\int_{0}^{1}\sqrt{t-t^2}\,\mathrm{d}t\\ &=\frac12\int_{-1}^{1}\sqrt{\frac14(1-u^2)}\,\mathrm{d}u;~~[u=2t-1]\\ &=\frac12\int_{0}^{1}\sqrt{1-u^2}\,\mathrm{d}u\\ &=\frac{\pi}{8}.\\ \end{align}$$

Esto nos da,

$$\mathcal{I}=\frac{\pi^2}{4\sqrt{2}}\,I_{1}+\frac{1}{\sqrt{2}}\,I_{2}=\frac{\pi^2}{4\sqrt{2}}\cdot\frac{\pi}{8}+\frac{1}{\sqrt{2}}\,I_{2}=\frac{\pi^3}{32\sqrt{2}}+\frac{1}{\sqrt{2}}\,I_{2}.$$

Aquí es la antiderivada $g{(z)}$:

$$\begin{align} g{(z)} &=\int\sqrt{\cos{\left(z\right)}-\cos^2{\left(z\right)}}\cdot\sin{\left(z\right)}\,\mathrm{d}z\\ &=\int\sqrt{\cos{\left(z\right)}-\cos^2{\left(z\right)}}\cdot\sin{\left(z\right)}\,\mathrm{d}z\\ &=-\int\sqrt{t-t^2}\,\mathrm{d}t;~~[t=\cos{z}]\\ &=-\int\sqrt{\frac14\left(1-(2t-1)^2\right)}\,\mathrm{d}t\\ &=-\frac14\int\sqrt{1-u^2}\,\mathrm{d}u;~~[u=2t-1]\\ &=-\frac18u\sqrt{1-u^2}-\frac18\arcsin{\left(u\right)}+\color{grey}{constant}\\ &=\frac14\left(1-2t\right)\sqrt{t-t^2}+\frac18\arcsin{\left(1-2t\right)}+\color{grey}{constant}\\ &=\frac14\left(1-2\cos{z}\right)\sqrt{\cos{z}-\cos^2{z}}+\frac18\arcsin{\left(1-2\cos{z}\right)}+\color{grey}{constant}.\\ \end{align}$$

La segunda integral de la $I_2$ es donde la mayoría de las dificultades se encuentran. El uso de la antiderivada $g{(z)}$ a integrar $I_2$ por partes, podemos reducir su evaluación constante, además de otras dos integrales:

$$\begin{align} I_{2} &=\int_{0}^{\frac{\pi}{2}}z^2\sqrt{\cos{\left(z\right)}-\cos^2{\left(z\right)}}\cdot\sin{\left(z\right)}\,\mathrm{d}z\\ &=\left[z^2\left(\frac14\left(1-2\cos{z}\right)\sqrt{\cos{z}-\cos^2{z}}+\frac18\arcsin{\left(1-2\cos{z}\right)}\right)\right]_{0}^{\frac{\pi}{2}}\\ &~~~~~ -\int_{0}^{\frac{\pi}{2}}2z\left(\frac14\left(1-2\cos{z}\right)\sqrt{\cos{z}-\cos^2{z}}+\frac18\arcsin{\left(1-2\cos{z}\right)}\right)\,\mathrm{d}z\\ &=\left[\frac{z^2}{4}\left(1-2\cos{z}\right)\sqrt{\cos{z}-\cos^2{z}}+\frac{z^2}{8}\arcsin{\left(1-2\cos{z}\right)}\right]_{0}^{\frac{\pi}{2}}\\ &~~~~~ +\int_{0}^{\frac{\pi}{2}}\left(\frac{z}{2}\left(2\cos{z}-1\right)\sqrt{\cos{z}-\cos^2{z}}+\frac{z}{4}\arcsin{\left(2\cos{z}-1\right)}\right)\,\mathrm{d}z\\ &=\left[\frac{z^2}{8}\arcsin{\left(1-2\cos{z}\right)}\right]_{0}^{\frac{\pi}{2}}+\frac12\int_{0}^{\frac{\pi}{2}}z\left(2\cos{z}-1\right)\sqrt{\cos{z}-\cos^2{z}}\,\mathrm{d}z\\ &~~~~~ +\frac14\int_{0}^{\frac{\pi}{2}}z\arcsin{\left(2\cos{z}-1\right)}\,\mathrm{d}z\\ &=\frac{\pi^3}{64}+\frac12\int_{0}^{\frac{\pi}{2}}z\left(2\cos{z}-1\right)\sqrt{\cos{z}-\cos^2{z}}\,\mathrm{d}z+\frac14\int_{0}^{\frac{\pi}{2}}z\arcsin{\left(2\cos{z}-1\right)}\,\mathrm{d}z\\ &=:\frac{\pi^3}{64}+\frac12\,J_{1}+\frac14\,J_{2}.\\ \end{align}$$

Esto produce:

$$\mathcal{I}=\frac{\pi^3}{32\sqrt{2}}+\frac{1}{\sqrt{2}}\,\left(\frac{\pi^3}{64}+\frac12\,J_{1}+\frac14\,J_{2}\right)=\frac{3\sqrt{2}\,\pi^3}{128}+\frac{\sqrt{2}}{4}\,J_{1}+\frac{\sqrt{2}}{8}\,J_{2}.$$

El primero de los dos restantes integrales parece tener el valor,

$$J_{1}=\frac{5\pi\ln{(2)}}{4}-\frac{7\pi}{8}\approx-0.026911.$$

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EDITAR:

Podemos demostrar esto a través de la integración por partes con el hecho de que la función trigonométrica $\left(2\cos{z}-1\right)\sqrt{\cos{z}-\cos^2{z}}$, al igual que el anterior, tiene una escuela primaria antiderivada también, que vamos a denotar por $h{(z)}$. Comenzamos por el uso de la tangente de la mitad de ángulo de sustitución de $t=\tan{\left(\frac{z}{2}\right)}$ a transformar la integral indefinida $h{(z)}$ y nuestra forma de trabajo a partir de allí:

$$\begin{align} h{(z)} &=\int\left(2\cos{z}-1\right)\sqrt{\cos{z}-\cos^2{z}}\,\mathrm{d}z;~~[-\frac{\pi}{2}<z<\frac{\pi}{2}]\\ &=\small{\int\left(2\left(\frac{1-t^2}{1+t^2}\right)-1\right)\sqrt{\left(\frac{1-t^2}{1+t^2}\right)-\left(\frac{1-t^2}{1+t^2}\right)^2}\cdot\frac{2\,\mathrm{d}t}{1+t^2}};~~[t=\tan{\left(\frac{z}{2}\right)}]\\ &=2\int\frac{1-3t^2}{(1+t^2)^2}\sqrt{\frac{2t^2(1-t^2)}{(1+t^2)^2}}\,\mathrm{d}t\\ &=2\sqrt{2}\int\frac{1-3t^2}{(1+t^2)^3}\cdot|t|\sqrt{1-t^2}\,\mathrm{d}t\\ &=2\sqrt{2}\int\frac{1-3t^2}{(1+t^2)^3}t\sqrt{1-t^2}\,\mathrm{d}t;~~[0\le t<1]\\ &=2\sqrt{2}\int\frac{1-3t^2}{(1+t^2)^3}\cdot\frac{t(1-t^2)}{\sqrt{1-t^2}}\,\mathrm{d}t\\ &=2\sqrt{2}\int\frac{(3t^2-1)(1-t^2)}{(1+t^2)^3}\cdot\frac{(-t)}{\sqrt{1-t^2}}\,\mathrm{d}t\\ &=2\sqrt{2}\int\left[-\frac{3}{1+t^2}+\frac{10}{(1+t^2)^2}-\frac{8}{(1+t^2)^3}\right]\cdot\frac{(-t)}{\sqrt{1-t^2}}\,\mathrm{d}t\\ &=2\sqrt{2}\int\left[-\frac{3}{2-u^2}+\frac{10}{(2-u^2)^2}-\frac{8}{(2-u^2)^3}\right]\,\mathrm{d}u;~~[u=\sqrt{1-t^2}]\\ &=4\int\left[-\frac{3}{2-2v^2}+\frac{10}{(2-2v^2)^2}-\frac{8}{(2-2v^2)^3}\right]\,\mathrm{d}v;~~{v=\frac{1}{\sqrt{2}}u}\\ &=2\int\left[-\frac{3}{1-v^2}+\frac{5}{(1-v^2)^2}-\frac{2}{(1-v^2)^3}\right]\,\mathrm{d}v\\ &=\int\left[\frac{6v^2}{(1-v^2)^2}-\frac{4v^2}{(1-v^2)^3}\right]\,\mathrm{d}v\\ &=\int\left[\frac{6v^2(1-v^2)}{(1-v^2)^3}-\frac{4v^2}{(1-v^2)^3}\right]\,\mathrm{d}v\\ &=\int\left(v-3v^3\right)\cdot\frac{2v}{(1-v^2)^3}\,\mathrm{d}v\\ &=\left[\frac{v-3v^3}{2(1-v^2)^2}\right]-\int\frac{1-9v^2}{2(1-v^2)^2}\,\mathrm{d}v\\ &=\frac{v-3v^3}{2(1-v^2)^2}+\int\frac{9v^2-1}{2(1-v^2)^2}\,\mathrm{d}v\\ &=\frac{v-3v^3}{2(1-v^2)^2}+\int\frac{4v^2}{(1-v^2)^2}\,\mathrm{d}v-\frac12\int\frac{\mathrm{d}v}{1-v^2}\\ &=\frac{v-3v^3}{2(1-v^2)^2}+\int(2v)\cdot\frac{2v}{(1-v^2)^2}\,\mathrm{d}v-\frac12\int\frac{\mathrm{d}v}{1-v^2}\\ &=\frac{v-3v^3}{2(1-v^2)^2}+\left[\frac{2v}{1-v^2}\right]-\int\frac{2}{1-v^2}\,\mathrm{d}v-\frac12\int\frac{\mathrm{d}v}{1-v^2}\\ &=\frac{v(5-7v^2)}{2(1-v^2)^2}-\frac52\int\frac{\mathrm{d}v}{1-v^2}\\ &=\frac{v(5-7v^2)}{2(1-v^2)^2}-\frac52\tanh^{-1}{\left(v\right)}+\color{grey}{constant}\\ &=\frac{u(10-7u^2)}{\sqrt{2}(2-u^2)^2}-\frac52\tanh^{-1}{\left(\frac{u}{\sqrt{2}}\right)}+\color{grey}{constant}\\ &=\frac{(3+7t^2)}{\sqrt{2}(1+t^2)^2}\sqrt{1-t^2}-\frac52\tanh^{-1}{\left(\frac{\sqrt{1-t^2}}{\sqrt{2}}\right)}+\color{grey}{constant}\\ &=\frac{(3+7t^2)}{\sqrt{2}(1+t^2)^{3/2}}\sqrt{\frac{1-t^2}{1+t^2}}-\frac52\tanh^{-1}{\left(\frac{\sqrt{1-t^2}}{\sqrt{2}}\right)}+\color{grey}{constant}\\ &=\frac{5-2\cos{\left(z\right)}}{\sqrt{2}\left|\sec{\left(\frac{z}{2}\right)}\right|}\sqrt{\cos{\left(z\right)}}-\frac52\tanh^{-1}{\left(\frac{\sqrt{1-\tan^2{\left(\frac{z}{2}\right)}}}{\sqrt{2}}\right)}+\color{grey}{constant}\\ &=\frac{5-2\cos{\left(z\right)}}{2}\sqrt{\cos{\left(z\right)}+\cos^2{\left(z\right)}}-\frac52\tanh^{-1}{\left(\sqrt{\frac{\cos{\left(z\right)}}{1+\cos{\left(z\right)}}}\right)}+\color{grey}{constant}.\\ \end{align}$$

El uso de la antiderivada $h{(z)}$ se encuentra por encima, podemos integrar a $J_1$ por partes (nota: el límite de términos desaparecer), a fin de reducir a un puramente trigonométricas integral, el cual puede ser manejado de la forma habitual por la tangente de la mitad de ángulo de sustitución, $\tan{\left(\frac{z}{2}\right)}=t$:

$$\begin{align} J_{1} &=\int_{0}^{\frac{\pi}{2}}z\left(2\cos{\left(z\right)}-1\right)\sqrt{\cos{\left(z\right)}-\cos^2{\left(z\right)}}\,\mathrm{d}z\\ &=\small{\left[z\left(\frac{5-2\cos{\left(z\right)}}{2}\sqrt{\cos{\left(z\right)}+\cos^2{\left(z\right)}}-\frac52\tanh^{-1}{\left(\sqrt{\frac{\cos{\left(z\right)}}{1+\cos{\left(z\right)}}}\right)}\right)\right]_{0}^{\frac{\pi}{2}}}\\ &~~~~~ \small{-\int_{0}^{\frac{\pi}{2}}\left[\frac{5-2\cos{\left(z\right)}}{2}\sqrt{\cos{\left(z\right)}+\cos^2{\left(z\right)}}-\frac52\tanh^{-1}{\left(\sqrt{\frac{\cos{\left(z\right)}}{1+\cos{\left(z\right)}}}\right)}\right]\,\mathrm{d}z}\\ &=\small{\frac{\pi}{2}\cdot0-0\cdot\left(\frac{3}{\sqrt{2}}-\frac52\tanh^{-1}{\left(\frac{1}{\sqrt{2}}\right)}\right)+\frac12\int_{0}^{\frac{\pi}{2}}\left(2\cos{\left(z\right)}-5\right)\sqrt{\cos{\left(z\right)}+\cos^2{\left(z\right)}}\,\mathrm{d}z}\\ &~~~~~ \small{+\frac52\int_{0}^{\frac{\pi}{2}}\tanh^{-1}{\left(\sqrt{\frac{\cos{\left(z\right)}}{1+\cos{\left(z\right)}}}\right)}\,\mathrm{d}z}\\ &=\small{\frac12\int_{0}^{\frac{\pi}{2}}\left(2\cos{\left(z\right)}-5\right)\sqrt{\cos{\left(z\right)}+\cos^2{\left(z\right)}}\,\mathrm{d}z+\frac52\int_{0}^{\frac{\pi}{2}}\tanh^{-1}{\left(\sqrt{\frac{\cos{\left(z\right)}}{1+\cos{\left(z\right)}}}\right)}\,\mathrm{d}z}\\ &=\small{\frac12\int_{0}^{1}\left[2\cdot\frac{1-t^2}{1+t^2}-5\right]\sqrt{\frac{2(1-t^2)}{(1+t^2)^2}}\cdot\frac{2}{1+t^2}\,\mathrm{d}t+\frac52\int_{0}^{1}\tanh^{-1}{\left(\frac{\sqrt{1-t^2}}{\sqrt{2}}\right)}\cdot\frac{2}{1+t^2}\,\mathrm{d}t}\\ &=-\sqrt{2}\int_{0}^{1}\frac{3+7t^2}{(1+t^2)^3}\sqrt{1-t^2}\,\mathrm{d}t+5\int_{0}^{1}\frac{\tanh^{-1}{\left(\frac{\sqrt{1-t^2}}{\sqrt{2}}\right)}}{1+t^2}\,\mathrm{d}t\\ &=-\sqrt{2}\left[\frac{7\pi}{8\sqrt{2}}\right]+5\left[\frac{\pi\ln{(2)}}{4}\right]\\ &=-\frac{7\pi}{8}+\frac{5\pi\ln{(2)}}{4}.\\ \end{align}$$

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De nuevo vamos a dividir el restante integral de la $J_2$ en más porciones manejables mediante integración por partes y sustituciones apropiadas.

$$\begin{align} J_{2} &=\int_{0}^{\frac{\pi}{2}}x\arcsin{\left(2\cos{\left(x\right)}-1\right)}\,\mathrm{d}x\\ &=-\int_{-1}^{1}\frac{\arccos{\left(\frac{1-y}{2}\right)}\,\arcsin{\left(y\right)}}{\sqrt{(3-y)(y+1)}}\,\mathrm{d}y;~~[y=1-2\cos{\left(x\right)}]\\ &=\int_{0}^{1}\frac{\arccos{\left(z\right)}\,\arcsin{\left(2z-1\right)}}{\sqrt{1-z^2}}\,\mathrm{d}z;~~[z=\frac{1-y}{2}]\\ &=-\frac{\pi^3}{16}+\frac12\int_{0}^{1}\frac{\arccos{\left(z\right)}^2}{\sqrt{z(1-z)}}\,\mathrm{d}z;~~[\text{I.B.P.}]\\ &=-\frac{\pi^3}{16}+\frac12\int_{0}^{1}\frac{\left[\frac{\pi}{2}-\arcsin{\left(z\right)}\right]^2}{\sqrt{z(1-z)}}\,\mathrm{d}z\\ &=-\frac{\pi^3}{16}+\frac12\int_{0}^{1}\frac{\left[\frac{\pi^2}{4}-\pi\arcsin{\left(z\right)}+\arcsin{\left(z\right)}^2\right]}{\sqrt{z(1-z)}}\,\mathrm{d}z\\ &=-\frac{\pi^3}{16}+\frac{\pi^2}{8}\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{z(1-z)}}-\frac{\pi}{2}\int_{0}^{1}\frac{\arcsin{\left(z\right)}}{\sqrt{z(1-z)}}\,\mathrm{d}z+\frac12\int_{0}^{1}\frac{\arcsin{\left(z\right)}^2}{\sqrt{z(1-z)}}\,\mathrm{d}z\\ &=+\frac{\pi^3}{16}-\frac{\pi}{2}\int_{0}^{1}\frac{\arcsin{\left(z\right)}}{\sqrt{z(1-z)}}\,\mathrm{d}z+\frac12\int_{0}^{1}\frac{\arcsin{\left(z\right)}^2}{\sqrt{z(1-z)}}\,\mathrm{d}z\\ &=:\frac{\pi^3}{16}-\pi\,K_{1}+K_{2}.\\ \end{align}$$

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Actualización con la justificación próxima...

$$K_{1}=\frac{\pi}{4}\,{_4F_3}{\left(\frac12,\frac12,\frac34,\frac54;\frac32,1,\frac32;1\right)}.$$

$$K_{2}=\frac{\pi^3}{48}-\frac{\pi}{2}\operatorname{Li}_{2}{\left(-\delta^{-2}\right)}+\pi\ln{\left(2\right)}\ln{\left(\frac{\delta}{2}\right)}+\pi\operatorname{Li}_{2}{\left(\delta^{-1}\right)}-\pi\operatorname{Li}_{2}{\left(\frac12\right)},$$

donde $\delta$ aquí denota la "plata ratio", $\delta:=1+\sqrt{2}$.