para

Question

Quiero mostrar

$$\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} = \frac{\pi}{2[a(a+1)]^\frac{1}{2}}$$

$a > 0$.

Probar varios métodos

1. Sustituciones para racionalizar
2. El famoso U = $\tan(\frac{\theta}{2})$
3. Multiplicando por conjugados

y otras técnicas de cálculo y aún así, no podemos demostrar la igualdad.

¿Alguien me puede dar una sugerencia de cómo debo empezar a trabajar esta integral?

Me siento muy avergonzado de que no puedo solucionar este problema.

Answer 1

Sugerencia Que $u=\tan(\theta),$ o $\theta=\arctan(u).$ entonces la integral se convierte

$$\int_{0}^{\infty} \frac{1}{a+\frac{u^2}{1+u^2}} \frac{1}{1+u^2} \ du= \int_{0}^{\infty} \frac{1}{(a+1)u^2+a} \ du,$$

¿Lo puede tomar desde aquí?

Answer 2

Multiplicar numerador y denominador por $\csc^2x$,

$$ \int \frac{\csc^2 \theta}{a \csc^2 \theta+1} \,d\theta$$

utilizar $\csc^2\theta = \cot^2\theta+1$ y sustituye $u\mapsto \cot \theta$,

$$ - \int \frac{1}{au^2 + a + 1}\,du $$

factor $1/(a+1)$,

$$ \frac{-1}{a+1} \int \frac{1}{\frac{au^2}{a+1}+1}\,du $$

sustituir $t\mapsto u\sqrt{\frac{a}{a+1}}$,

$$ -\frac 1 a \sqrt{\frac{a}{a+1}} \int \frac{1}{t^2+1}\, dt $$

usar la derivada de arctan,

$$ -\frac 1 a \sqrt{\frac{a}{a+1}} \arctan(t) + c $$

y por último en llegar,

$$ \frac{1}{\sqrt{a(a+1)}} \cdot \arctan\left( \frac{\tan \theta}{\sqrt{a/(a+1)}}\right) + c $$

Ahora evaluar en los extremos y debe tener el lado derecho de su identidad.

Answer 3

\begin{align} \int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + > \sin^2(\theta)} &= \frac{\pi}{2[a(a+1)]^\frac{1}{2}} \end {Alinee el}

\begin{align} I= \int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} &= \int_{0}^{\frac{\pi}{2}}\frac{d\theta}{(\sqrt{a+1} + \cos\theta) (\sqrt{a+1} - \cos\theta} \\ &=\frac1{2\sqrt{a+1}} \left( \int_0^{\tfrac\pi2} \frac{d\theta}{\sqrt{a+1} + \cos\theta} + \int_0^{\tfrac\pi2} \frac{d\theta}{\sqrt{a+1} - \cos\theta} \right) . \end {Alinee el}

Ahora podemos utilizar un integral de tabla

\begin{align} \int \frac{dx}{u+v\cos x} &= \frac2{\sqrt{u^2-v^2}} \arctan\left( \frac{(u-v)\tan\tfrac{x}2}{\sqrt{u^2-v^2}} \right) ,\quad u^2>v^2 \end {Alinee el}

$u=\sqrt{a+1}$, $v=\pm 1$.

\begin{align} I&= \frac1{2\,u} \frac2{\sqrt{u^2-1}} \left( \arctan\left( \sqrt{\frac{u-1}{u+1}}\tan\tfrac\pi4 \right) + \arctan\left( \sqrt{\frac{u+1}{u-1}}\tan\tfrac\pi4 \right) \right) ,\\ &= \frac1{u\,\sqrt{u^2-1}} \left( \arctan\left( \sqrt{\frac{u-1}{u+1}} \right) + \arctan\left( \sqrt{\frac{u+1}{u-1}} \right) \right) . \end {Alinee el}

Y desde $\arctan x+\arctan \tfrac1x=\tfrac\pi2$ $x>0$, tenemos

\begin{align} I&= \frac{\pi}{2\,\sqrt{a+1}\,\sqrt{a}} . \end {Alinee el}

Answer 4

Otro método es el siguiente. Utilizando $2\sin^{2}t = 1 - \cos 2t$ podemos ver que $$\int_{0}^{\pi/2}\frac{dt}{a + \sin^{2}t} = 2\int_{0}^{\pi/2}\frac{dt}{2a + 1 - \cos 2t}$$ and putting $ 2t = z $ we get the integral as $$\int_{0}^{\pi}\frac{dz}{2a + 1 - \cos z}$$ which is equal to $$\frac{\pi}{\sqrt{(2a + 1)^{2} - 1^{2}}} = \frac{\pi}{2\sqrt{a(a+1)}}$$ The above integral is a special case of the general formula $$\int_{0}^{\pi}\frac{dx}{A + B\cos x} = \int_{0}^{\pi}\frac{dx}{A - B\cos x} = \frac{\pi}{\sqrt{A^{2} - B^{2}}}$$ which holds if $A > | B | $. The general formula can be established using the substitution $% $ $(A + B\cos x)(A - B\cos y)= A^{2} - B^{2}$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\pi/2}{\dd\theta \over a + \sin^{2}\pars{\theta}} & = \int_{0}^{\pi/2}{\sec^{2}\pars{\theta}\,\dd\theta \over a\sec^{2}\pars{\theta} + \tan^{2}\pars{\theta}} = \int_{0}^{\pi/2}{\sec^{2}\pars{\theta}\,\dd\theta \over \pars{a + 1}\tan^{2}\pars{\theta} + a} \\[5mm] & = {1 \over a}\,\root{a \over a + 1}\int_{0}^{\pi/2}{% \root{\pars{a + 1}/a}\sec^{2}\pars{\theta}\,\dd\theta \over \bracks{\root{\pars{a + 1}/a}\tan\pars{\theta}}^{\,2} + 1} \\[5mm] & \stackrel{x\ \equiv\ \root{\pars{a + 1}/a}\tan\pars{\theta}}{=}\,\,\, {1 \over \root{a\pars{a + 1}}}\ \underbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}} _{\ds{=\ {\pi \over 2}}}\ =\ \bbx{\pi \over 2\,\bracks{a\pars{a + 1}}^{1/2}} \end{align}