• Matéria: Matemática
  • Autor: jessiorg
  • Perguntado 8 anos atrás

Qual é a integral da raiz de 1+x^2 dx?

Respostas

respondido por: FibonacciTH
22
Dada a integral:

\mathsf{\int \:\sqrt{1+x^2}dx}

Iremos calcular a integral indefinida através do método da substituição trigonométrica. Lembrando da semelhança trigonométrica abaixo:

\mathsf{sec^2\theta =tg^2\theta \:+1}

Considerando que:

\mathsf{sec^2\theta =1+x^2}\\\\\mathsf{1+tg^2\theta =1+x^2}\\\\\mathsf{tg^2\theta =x^2}\\\\\mathsf{tg\:\theta =x}

Calculando a derivada de \mathsf{x} em função de \mathsf{\theta }:

\mathsf{dx=sec^2\theta \:d\theta }

Substituindo na integral teremos:

\mathsf{\int \:\sqrt{sec^2\theta }\cdot sec^2\:d\theta }\\\mathsf{\int \:sec\theta \cdot sec^2\theta \:d\theta }\\\mathsf{\int \:sec^3\theta \:d\theta }

Devendo lembrar da propriedade:

\mathsf{\int \:sec^n\left(x\right)\:dx=\dfrac{1}{n-1}\cdot \left(sec^{n-2}\left(x\right)\cdot tg\left(x\right)+n-2\int \:sec^{n-2}\left(x\right)dx\right)}

Aplicando:

\mathsf{\int \sec^3\left(\theta\right)\:dx=\dfrac{1}{3-1}\cdot \left(sec^{3-2}\left(\theta\right)\cdot tg\left(\theta\right)+3-2\int sec^{3-2}\left(\theta \right)d\theta\right)}\\\\\\\mathsf{\int sec^3\left(\theta \right)\:dx=\dfrac{1}{2}\cdot \left(sec\left(\theta \right)\cdot tg\left(\theta \right)+\int sec\left(\theta \right)d\theta \right)}
\mathsf{\int sec^3\left(\theta \right)\:dx=\dfrac{1}{2}\cdot \left(sec\left(\theta \right)\cdot tg\left(\theta \right)+ln\left[tg\left(\theta \right)+sec\left(\theta \right)\right]\right)+C}\\\\\\\mathsf{\int sec^3\left(\theta \right)\:dx=\dfrac{1}{2}\cdot \left(\left[\sqrt{1+x^2}\right]\cdot x+ln\left[x+\sqrt{1+x^2}\right]\right)+C}\\\\\\\mathsf{\int sec^3\left(\theta \right)\:dx=\dfrac{1}{2}\cdot \left(x\sqrt{1+x^2}+ln\left[x+\sqrt{1+x^2}\right]\right)+C}
\mathsf{\int sec^3\left(\theta \right)\:dx=\left(\dfrac{x\sqrt{1+x^2}}{2}+\dfrac{ln\left(x+\sqrt{1+x^2}\right)}{2}\right)+C}\\\\\\\mathsf{\mathsf{\int sec^3\left(\theta \right)\:dx=\left(\dfrac{x\sqrt{1+x^2}+ln\left(x+\sqrt{1+x^2}\right)}{2}\right)+C}}

= = = = =

\boxed{\mathsf{Resposta:\int \sqrt{1+x^2}\:dx=\left(\dfrac{x\sqrt{1+x^2}+ln\left(x+\sqrt{1+x^2}\right)}{2}\right)+C}}
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