cálculo de $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ y $\int\frac{1}{\sin^5 x+\cos^5x}dx$

Question

Resolver el siguiente indefinido integrales:

$$ \begin{align} &(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\ &(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx \end{align} $$

Mi Intento por $(1)$:

$$ \begin{align} I &= \int\frac{1}{\sin^3 x+\cos ^3 x}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^2 x+\cos ^2 x-\sin x \cos x\right)}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}\;dx\\ &= \frac{1}{3}\int \left(\frac{2}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x+\cos x \right)}{\left(1-\sin x\cos x\right)}\right)\;dx\\ &= \frac{2}{3}\int\frac{1}{\sin x+\cos x}\;dx + \frac{1}{3}\int\frac{\sin x+\cos x}{1-\sin x\cos x}\;dx \end{align} $$

El uso de las identidades

$$ \sen x = \frac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}},\;\cos x = \frac{1-\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}} $$

podemos transformar la integral a

$$I = \frac{1}{3}\int\frac{\left(\tan \frac{x}{2}\right)^{'}}{1-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}}\;dx+\frac{2}{3}\int\frac{\left(\sin x- \cos x\right)^{'}}{1+(\sen x-\cos x)^2}\;dx $$

La integral es fácil de calcular a partir de aquí.

Mi Intento por $(2)$:

$$ \begin{align} J &= \int\frac{1}{\sin^5 x+\cos ^5 x}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^4 x -\sin^3 x\cos x+\sin^2 x\cos^2 x-\sin x\cos^3 x+\cos^4 x\right)}\;dx\\ &= \int\frac{1}{(\sin x+\cos x)(1-2\sin^2 x\cos^2 x-\sin x\cos x+\sin^2 x\cos^2 x)}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x-\left(\sin x\cos x\right)^2\right)}\;dx \end{align} $$

Cómo puedo solucionar $(2)$ a partir de este punto?

Answer 1

Dadas $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx$ $

Primero simplificamos $$\sin^5 x+\cos^5 x = \left(\sin^2 x+\cos^2 x\right)\cdot \left(\sin^3 x+\cos^3 x\right) - \sin ^2x\cdot \cos ^2x\left(\sin x+\cos x\right)$ $

$$\displaystyle \sin^5 x+\cos^5 x= (\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)$$

Tan Integral es $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx $ $

$$\displaystyle = \int\frac{1}{(\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$

$$\displaystyle = \int \frac{(\sin x+\cos x)}{(\sin x+\cos x)^2\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$

$$\displaystyle = \int \frac{(\sin x+\cos x)}{(1+\sin 2x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$

Let $$(\sin x-\cos x) = t\;,$$ Then $$(\cos +\sin x)dx = dt$$ and $$(1-\sin 2x) = t^2\Rightarrow (1+\sin 2x) = (2-t^2)$$

Tan Integral convierten en $$\displaystyle = 4\int\frac{1}{(2-t^2)\cdot(5-t^4)}dt = 4\int\frac{1}{(t^2-2)\cdot (t^2-\sqrt{5})\cdot (t^2+\sqrt{5})}dt$ $

Ahora usando la fracción parcial, obtenemos

$$\displaystyle = 4\int \left[\frac{1}{2-t^2}+\frac{1}{(2-\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}-t^2)}+\frac{1}{(2+\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}+t^2)}\right]dt$$

$$ = \displaystyle \sqrt{2}\ln \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{(2-\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \ln \left|\frac{5^{\frac{1}{4}}+t}{5^{\frac{1}{4}}-t}\right|+\frac{2}{(2+\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \tan^{-1}\left(\frac{t}{5^{\frac{1}{4}}}\right)+\mathbb{C}$$

donde $$t=(\sin x-\cos x)$ $

Answer 2

$\newcommand{\+}{^{\daga}}% \newcommand{\ángulos}[1]{\left\langle #1 \right\rangle}% \newcommand{\llaves}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \cima {= \cima \vphantom{\enorme}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\mitad}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\a la derecha\vert\,}% \newcommand{\cy}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert}$ $\large\tt\mbox{Just a hint:}$ Escribir $$ \int{\cos\pars{x}\,\dd x \\cos\pars{x}\sin^{3}\pars{x} + \cos^{4}\pars{x}} = \int{\dd z \\raíz{1 - z^{2}}z^{3} + \bracks{1 - z^{2}}^{2}} \quad\mbox{con}\quad z \equiv \sin\pars{x} $$ El uso de una sustitución de Euler: $\root{1 - z^{2}} \equiv t + \ic z$ que los rendimientos de $1 - z^{2} = t^{2} + 2t\ic z - z^{2}$ tal que $\ds{z = {1 - t^{2} \over 2t\ic}}$: \begin{align} \root{1 - z^{2}}&=t + {1 - t^{2} \over 2t} = {1 + t^{2} \over 2t} \\[3mm] \dd z&= {\pars{-2t}\pars{2t\ic} - \pars{2\ic}\pars{1 - t^{2}} \over -4t^{2}}\,\dd t = \ic\,{t^{2} + 1 \over 2t^{2}}\,\dd t \end{align} \begin{align} \int&=\int{1 \over \bracks{\pars{1 + t^{2}}/2t}\bracks{\pars{1 - t^{2}}/2t}^{3}\pars{-1/\ic} + \bracks{\pars{1 + t^{2}}/2t}^{4}} \,\ic\,{t^{2} + 1 \over 2t^{2}}\,\dd t \\[3mm]&=-8\int{t^{2} \over -\pars{1 - t^{2}}^{3} + \ic\pars{1 + t^{2}}^{3}}\,\dd t \end{align}

Answer 3

No estoy seguro cómo se puede seguir cualquiera (el segundo término en el denominador se puede expresar como $1-\sin(2 x)/2 - \sin^2(2x)/4,$ pero no conozco ninguna fórmula de doble ángulo $\sin x + \cos x.$ el enfoque más sencillo para su integral es utilizar la substitución de $u = \tan \frac{x}2$ temido, que reduce el integral de una función racional integral...