Question

a integral de raiz quadrada de x^2 + a^2/x^4

Answer 1

É dada a seguinte integral:

$\displaystyle \int \frac{\sqrt{a^2+x^2}}{x^4} \, \, dx$

Podemos utilizar a seguinte substituição trigonométrica:

$x = a \tan \theta \\ \\ dx = a \sec^2 \theta \, \, d \theta$

Daí temos:

$\displaystyle \int \frac{\sqrt{a^2+(a \tan \theta)^2}}{(a \tan \theta)^4} \cdot a \sec^2 \theta \, \, d \theta \\\\\\ \int \frac{\sqrt{a^2+a^2 \tan^2 \theta}}{a^4 \tan^4 \theta} \cdot a \sec^2 \theta \, \, d \theta \\\\\\ \int \frac{\sqrt{a^2 \cdot (1+ \tan^2 \theta)}}{a^3 \tan^4 \theta} \cdot \sec^2 \theta \, \, d \theta$

De acordo com a seguinte identidade trigonométrica:

$\sec^2 \theta - \tan^2 \theta = 1 \\ \\ \sec^2 \theta = 1 + \tan^2 \theta$

Temos:

$\displaystyle \int \frac{\sqrt{a^2 \sec^2 \theta}}{a^3 \tan^4 \theta} \cdot \sec^2 \theta \, \, d \theta \\\\\\ \int \frac{a \sec \theta}{a^3 \tan^4 \theta} \cdot \sec^2 \theta \, \, d \theta \\\\\\ \int \frac{ \sec^3 \theta}{a^2 \tan^4 \theta} \, \, d \theta \\\\\\ \frac{1}{a^2} \cdot \int \frac{\sec^3 \theta}{\tan^4 \theta} \, \, d \theta$

De acordo com as identidades trigonométricas:

$\displaystyle \sec \theta = \frac{1}{\cos \theta} \\\\\\ \tan \theta = \frac{\sin \theta}{\cos \theta}$

Temos:

$\displaystyle \frac{1}{a^2} \cdot \int \frac{\displaystyle \frac{1}{\cos^3 \theta}}{\displaystyle \frac{\sin^4 \theta}{\cos^4 \theta}} \, \, d \theta \\\\\\ \frac{1}{a^2} \cdot \int \frac{1}{\cos^3 \theta} \cdot \frac{\cos^4 \theta}{\sin^4 \theta} \, \, d \theta \\\\\\ \frac{1}{a^2} \cdot \int \frac{\cos \theta}{\sin^4 \theta} \, \, d \theta \\\\\\ \frac{1}{a^2} \cdot \int \frac{1}{\sin^4 \theta} \cdot \cos \theta \, \, d \theta$

Fazendo:

$u = \sin \theta \\\\ du = \cos \theta \, \, d \theta$

Temos:

$\displaystyle \frac{1}{a^2} \cdot \int \frac{1}{u^4} \, \, du \\\\\\ \frac{1}{a^2} \cdot \bigg( -\frac{1}{3u^3} \bigg) + C \\\\\\ -\frac{1}{3a^2 \sin^3 \theta} + C$

De acordo com a premissa:

$\displaystyle x = a \tan \theta \\\\\\ \tan \theta = \frac{x}{a}$

Podemos comparar as duas fórmulas que seguem e imaginar um triângulo reto:

$\displaystyle \tan \theta = \frac{x}{a} \\\\\\ \tan \theta = \frac{\mathsf{cateto \, \, oposto}}{\mathsf{cateto \, \, adjacente}}$

E esse triângulo reto terá o cateto oposto sendo igual a x e o cateto adjacente sendo igual ao a, e a hipotenusa h é dada pelo Teorema de Pitágoras:

$h^2 = a^2+x^2 \\\\ h = \sqrt{a^2+x^2}$

Com os valores em mãos do cateto oposto, adjacente e da hipotenusa, podemos finalmente encontrar uma expressão para sin³ θ :

$\displaystyle \sin \theta = \frac{\mathsf{cateto \, \, oposto}}{\mathsf{hipotenusa}} \\\\\\ \sin \theta = \frac{x}{\sqrt{a^2+x^2}} \\\\\\ \sin^3 \theta = \bigg( \frac{x}{\sqrt{a^2+x^2}} \bigg)^3 \\\\\\ \sin^3 \theta = \frac{x^3}{(a^2+x^2)^{3/2}}$

Daí temos a resposta final sendo:

$\displaystyle - \frac{1}{3a^2 \sin^3 \theta} + C \\\\\\\\ - \frac{1}{\displaystyle 3a^2 \cdot \frac{x^3}{(a^2+x^2)^{3/2}} } + C \\\\\\\\ -\frac{1}{\displaystyle \frac{3a^2x^3}{(a^2+x^2)^{3/2}}} + C \\\\\\\\ \boxed{ \boxed{ -\frac{(a^2+x^2)^{3/2}}{3a^2x^3} + C } }$