Calcule a integral indefinida por inspeção:.... Pergunta de ideia deLukyo

November 2019 1 137 Report

Calcule a integral indefinida por inspeção:

$\displaystyle\int\sec x\,\mathrm{tg\,}x\,(1+2\sec^2 x)e^{\sec^2 x}\,dx$

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Niiya

Verified answer

$\displaystyle\int\mathsf{sec\,x\,tg\,x\,(1+2sec^{2}x)\,e^{sec^{2}x}\,dx=\int(1+2sec^{2}x)\,e^{sec^{2}x}\,sec\,x\,tg\,x\,dx}$

Fazendo a substituição $\mathsf{u=sec\,x}$ , temos $\mathsf{du=sec\,x\,tg\,x\,dx}$ :

$\displaystyle\int\mathsf{sec\,x\,tg\,x\,(1+2sec^{2}x)\,e^{sec^{2}x}\,dx=\int(1+2u^{2})\,e^{u^{2}}\,du}$
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Considere a sequência de funções $\mathsf{f_{n}(u)=u^{n}e^{u^{2}}}$ . Para cada $\mathsf{n\in\mathbb{N}}$ , temos $\mathsf{f_{n}'(u)=n\,u^{n-1}\,e^{u^{2}}+u^{n}\,e^{u^{2}}2u=(n+2u^{2})\,u^{n-1}\,e^{u^{2}}}$

Note que, para $\mathsf{n=1}$ , temos $\mathsf{f_{1}(u)=u\,e^{u^{2}}}$ e $\mathsf{f_{1}'(u)=(1+2u^{2})\,e^{u^{2}}}$ .

Como $\mathsf{\int f_{1}'(u)\,du=f_{1}(u)+c}$ , temos

$\displaystyle\int\mathsf{sec\,x\,tg\,x\,(1+2sec^{2}x)\,e^{sec^{2}x}\,dx=\int(1+2u^{2})\,e^{u^{2}}\,du}\\\\\\\int\mathsf{sec\,x\,tg\,x\,(1+2sec^{2}x)\,e^{sec^{2}x}\,dx=\int f_{1}'(u)\,du}\\\\\\\int\mathsf{sec\,x\,tg\,x\,(1+2sec^{2}x)\,e^{sec^{2}x}\,dx=f_{1}(u)+c}\\\\\\\int\mathsf{sec\,x\,tg\,x\,(1+2sec^{2}x)\,e^{sec^{2}x}\,dx=u\,e^{u^{2}}+c}$

Como $\mathsf{u=sec\,x}$ ,

$\boxed{\boxed{\int\mathsf{sec\,x\,tg\,x\,(1+2sec^{2}x)\,e^{sec^{2}x}\,dx=sec\,x\,e^{sec^{2}x}+c}}}$

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Lukyo Excelente. Muito obrigado! :)