Question

Calcule a integral indefinida por inspeção:

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

Answer 1

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

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

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

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 $\mathsf{u=sec\,x}$,

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