Question

por substituição trigonometrica qual é a integral de x^3 dx/ raiz de x^2-9? gostaria de saber, por favor.

Answer 1

$\mathrm{\int\dfrac{x^3}{\sqrt{x^2-9}}\ dx\ \to\ \boxed{\mathrm{x=3\sec{\theta}}}}\\\\\\ \mathrm{\int\dfrac{(3\sec{\theta})^3}{\sqrt{(3\sec{\theta})^2-9}}\ d\theta=\int\dfrac{27\sec^3{\theta}}{\sqrt{9\sec^2{\theta}-9}}\ d\theta=\int\dfrac{27\sec^3{\theta}}{3\sqrt{\sec^2{\theta}-1}}\ d\theta=}\\\\\\ \mathrm{=9\int\dfrac{\sec^3{\theta}}{\sqrt{\tan^2{\theta}}}\ d\theta=9\int\dfrac{\frac{1}{\cos^3{\theta}}}{\frac{\sin{\theta}}{\cos{\theta}}}=9\int\csc{\theta}\sec^2{\theta}\ d\theta=}$

$\mathrm{=9\int\csc{\theta}(1+\tan^2{\theta})\ d\theta=9\int\sec{\theta}\tan{\theta}+\csc{\theta}\ d\theta=}\\\\\\ \mathrm{=9\bigg(\int\sec{\theta}\tan{\theta}\ d\theta+\int\csc{\theta}\ d\theta\bigg)=9\bigg(\sec{\theta}-\ln{|\csc{\theta}+\cot{\theta}|}\bigg)=}\\\\\\ \mathrm{=9\bigg(\sec{\theta}-\ln{\bigg|\dfrac{1+\cos{\theta}}{\sin{\theta}}\bigg|}\bigg)=9\bigg(\sec{\theta}-\ln{\bigg|\dfrac{1+\cos{\theta}}{\sqrt{(1+\cos{\theta})(1-\cos{\theta})}\bigg|}}\bigg)}$

$\mathrm{=9\bigg(\sec{\theta}-\ln\bigg|{\sqrt{\dfrac{1+\cos{\theta}}{1-\cos{\theta}}}}\bigg|\bigg)=9\bigg(\sec{\theta}-\dfrac{1}{2}\ln\bigg|{\dfrac{1+\cos{\theta}}{1-\cos{\theta}}}}\bigg|\bigg)}}}$

$\mathrm{\Rightarrow Como\ x=3\sec{\theta}\ \to\ \sec{\theta}=\dfrac{x}{3}\ \ \bigg\| \ \ \cos{\theta}=\dfrac{3}{x}}}\\\\\\ \mathrm{9\bigg(\dfrac{x}{3}-\dfrac{1}{2}\ln{\bigg|\dfrac{1+\frac{3}{x}}{1-\frac{3}{x}}\bigg|}\bigg)=\boxed{\mathbf{9\bigg(\dfrac{x}{3}-\ln{\bigg|\dfrac{x+3}{x-3}\bigg|}\bigg)}}}}$

Answer 2

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador https://brainly.com.br/tarefa/10491561

                                                                                                                               

$\displaystyle\sf\int\dfrac{x^3}{\sqrt{x^2-9}}dx\\\sf fac_{\!\!,}a~x=3sec(\theta)\implies dx=3sec(\theta)tg(\theta)d\theta\\\sf x^3=27sec^3(\theta)\\\sf\sqrt{x^2-9}=\sqrt{9sec^2(\theta)-9}=\sqrt{9\cdot(sec^2(\theta)-1)}\\\sf\sqrt{x^2-9}=\sqrt{9tg^2(\theta)}=3tg(\theta)$

$\displaystyle\sf\int\dfrac{x^3}{\sqrt{x^2-9}}dx=\int\dfrac{27sec^3(\theta)}{\backslash\!\!\!3\diagup\!\!\!tg(\theta)}\backslash\!\!\!3sec(\theta)\diagup\!\!\!tg(\theta)d\theta=27\int sec^4 d\theta\\\displaystyle\sf27\int sec^4(\theta)d\theta=27\int sec^2(\theta)\cdot sec^2(\theta)d\theta=27\int(1+tg^2(\theta))sec^2(\theta)d\theta\\\sf fac_{\!\!,}a ~t=tg(\theta)\implies dt=sec^2(\theta)d\theta\\\displaystyle\sf27\int(1+tg^2(\theta))sec^2(\theta)d\theta=27\int(1+t^2)dt=27t+9t^3+k$

$\displaystyle\sf27\int sec^4(\theta)d\theta=27tg(\theta)+9tg^3(\theta)+k\\\sf tg(\theta)=\dfrac{\sqrt{x^2-9}}{3}\\\sf 27tg(\theta)=\backslash\!\!\!27\dfrac{\sqrt{x^2-9}}{\backslash\!\!\!3}=9\sqrt{x^2-9}\\\sf 9tg^3(\theta)=9\left(\dfrac{\sqrt{x^2-9}}{3}\right)^3=\backslash\!\!\!9\dfrac{\sqrt{x^2-9}^3}{\backslash\!\!\!9\cdot3}\\\boxed{\displaystyle\sf\int\dfrac{x^3}{\sqrt{x^2-9}}~dx=9\sqrt{x^2-9}+\dfrac{(\sqrt{x^2-9})^3}{3}+k}$