Question

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

Answer 1

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:

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

= = = = =

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