Question

Preciso entender, por favor$\int\limits^a_b {} \, dx / x^{2} * \sqrt{ x^{2} +25}$

Answer 1

Vamos avaliar a integral indefinida abaixo

$\displaystyle\int\dfrac{1}{x^{2}\sqrt{x^{2}+25}}\,dx$

Fazendo $x=5\,tg(\theta),~-\frac{\pi}{2}~\textless~\theta~\textless~\frac{\pi}{2}$, temos $dx=5\,sec^{2}(\theta)\,d\theta$

Então:

$\displaystyle\int\frac{1}{x^{2}\sqrt{x^{2}+25}}\,dx=\int\dfrac{1}{[5\,tg(\theta)]^{2}\sqrt{[5\,tg(\theta)]^{2}+25}}5\,sec^{2}(\theta)\,d\theta\\\\\\=\int\frac{1}{5\,tg^{2}(\theta)\sqrt{25\,tg^{2}(\theta)+25}}sec^{2}(\theta)\,d\theta\\\\\\=\int\dfrac{1}{5\,tg^{2}(\theta)\sqrt{25\cdot[tg^{2}(\theta)+1]}}sec^{2}(\theta)\,d\theta\\\\\\=\int\dfrac{sec^{2}(\theta)}{5\,tg^{2}(\theta)\cdot5\sqrt{sec^{2}(\theta)}}\,d\theta\\\\\\=\dfrac{1}{25}\int\dfrac{sec^{2}(\theta)}{tg^{2}(\theta)\cdot sec(\theta)}\,d\theta$

$=\dfrac{1}{25}\displaystyle\int\dfrac{sec(\theta)}{tg^{2}(\theta)}\,d\theta=\dfrac{1}{25}\int\dfrac{1}{cos(\theta)}\cdot cotg^{2}(\theta)\,d\theta\\\\\\=\dfrac{1}{25}\int\dfrac{1}{cos(\theta)}\dfrac{cos^{2}(\theta)}{sen^{2}(\theta)}\,d\theta=\dfrac{1}{25}\int\dfrac{cos(\theta)}{sen^{2}(\theta)}\,d\theta$

Se fizermos $u=sen(\theta)$, temos $du=cos(\theta)\,d\theta$

$\dfrac{1}{25}\displaystyle\int\dfrac{cos(\theta)}{sen^{2}(\theta)}\,d\theta=\dfrac{1}{25}\int\dfrac{1}{u^{2}}\,du=\dfrac{1}{25}\int u^{-2}\,du\\\\\\=\dfrac{1}{25}\dfrac{u^{-1}}{(-1)}=-\dfrac{1}{25}\dfrac{1}{u}=-\dfrac{1}{25\,sen(\theta)}+C$

Agora, precisamos voltar para a variável x

$x=5\,tg(\theta)~~\rightarrow~~tg(\theta)=\dfrac{x}{5}~~\rightarrow~~cotg(\theta)=\dfrac{5}{x}$

Elevando os dois lados ao quadrado:

$cotg^{2}(\theta)=\dfrac{25}{x^{2}}\\\\\\cotg^{2}(\theta)+1=\dfrac{25}{x^{2}}+1=\dfrac{25+x^{2}}{x^{2}}$

Mas sabemos que $cotg^{2}(\theta)+1=cossec^{2}(\theta)=\dfrac{1}{sen^{2}(\theta)}$. Então:

$\dfrac{1}{sen^{2}(\theta)}=\dfrac{25+x^{2}}{x^{2}}~~\Leftrightarrow~~sen^{2}(\theta)=\dfrac{x^{2}}{25+x^{2}}$

Aplicando raiz quadrada nos dois lados da igualdade:

$|sen(\theta)|=\sqrt{\dfrac{x^{2}}{25+x^{2}}}$

Como $\theta\in(-\frac{\pi}{2},\frac{\pi}{2}),~sen(\theta)~\textgreater~0$, daí

$|sen(\theta)|=sen(\theta)=\sqrt{\dfrac{x^{2}}{x^{2}+25}}$

Portanto:

$\displaystyle\int\dfrac{dx}{x^{2}\sqrt{x^{2}+25}}=-\dfrac{1}{25}\dfrac{1}{sen(\theta)}+C=-\dfrac{1}{25}\sqrt{\dfrac{25+x^{2}}{x^{2}}}+C$
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Finalmente:

$\displaystyle\int\limits_{b}^{a}\dfrac{dx}{x^{2}\sqrt{x^{2}+25}}=\bigg[\int\dfrac{dx}{x^{2}\sqrt{x^{2}+25}}\bigg]_{b}^{a}=-\dfrac{1}{25}\bigg[\sqrt{\dfrac{25+x^{2}}{x^{2}}}\bigg]_{b}^{a}\\\\\\\therefore~~\boxed{\boxed{\int\limits_{b}^{a}\dfrac{dx}{x^{2}\sqrt{x^{2}+25}}=\dfrac{1}{25}\bigg[\sqrt{\dfrac{25+b^{2}}{b^{2}}}-\sqrt{\dfrac{25+a^{2}}{a^{2}}}\bigg]}}$