Matemática, perguntado por Lukyo, 1 ano atrás

Calcule a integral indefinida por inspeção:

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

Soluções para a tarefa

Respondido por Niiya
4
\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}}}\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}}}

Lukyo: Excelente. Muito obrigado! :)
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