Question

Calcule a integral por partes

Answer 1

$I=\displaystyle\int\!x\sec^2 x\,dx$


Método de integração por partes:

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


Faça a seguinte substituição:

$\cos x=w~~\Rightarrow~~-\mathrm{sen\,}x\,dx=dw$


e a integral $\mathbf{(i)}$ fica

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


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \:x\cdot~sec ^2\left(x\right)dx\\\\\\=xtan \left(x\right)-\int tan \left(x\right)dx\\\\\\=xtan \left(x\right)-\left(-ln \left|cos \left(x\right)\right|\right)\\\\\\=xtan \left(x\right)+ln \left|cos \left(x\right)\right|\\\\\\\to \boxed{\sf =xtan \left(x\right)+ln \left|cos \left(x\right)\right|+C}$