Question

integral(x-1)sec²x dx

Answer 1

$I=\displaystyle\int\!(x-1)\sec^2 x\,dx$


Método de integração por partes:

$\begin{array}{lcl} u=x-1=&~\Rightarrow~&du=dx\\\\ dv=\sec^2 x\,dx&~\Leftarrow~&v=\mathrm{tg\,}x \end{array}\\\\\\\\ \displaystyle\int\!u\,dv=uv-\int\!v\,du\\\\\\ \int\!(x-1)\sec^2 x\,dx=(x-1)\,\mathrm{tg\,}x-\int\!\mathrm{tg\,}x\,dx\\\\\\ \int\!(x-1)\sec^2 x\,dx=(x-1)\,\mathrm{tg\,}x-\int\!\frac{\mathrm{sen\,}x}{\cos x}\,dx$

$\displaystyle\int\!(x-1)\sec^2 x\,dx=(x-1)\,\mathrm{tg\,}x+\int\!\frac{1}{\cos x}\cdot (-\mathrm{sen\,}x)\,dx\\\\\\ \int\!(x-1)\sec^2 x\,dx=(x-1)\,\mathrm{tg\,}x+\int\!\frac{1}{w}\,dw\qquad(w=\cos x)\\\\\\ \int\!(x-1)\sec^2 x\,dx=(x-1)\,\mathrm{tg\,}x+\mathrm{\ell n}\left|w\right|+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!(x-1)\sec^2 x\,dx=(x-1)\,\mathrm{tg\,}x+\mathrm{\ell n}\left|\cos x\right|+C \end{array}}$


Dúvidas? Comente.


Bons estudos! :-)