Question

integral por partes de arcsen x dx

Answer 1

$\displaystyle\mathtt{\int\!arcsen\,x\,dx}$


Método de integração por partes:

$\begin{array}{lcl} \mathtt{u=arcsen\,x}&~\Rightarrow~&\mathtt{du=\dfrac{1}{\sqrt{1-x^2}}\,dx}\\\\ \mathtt{dv=dx}&~\Leftarrow~&\mathtt{v=x} \end{array}$


$\displaystyle\mathtt{\int\!u\,dv=uv-\int\!v\,du}\\\\\\ \mathtt{\int\!arcsen\,x\,dx=arcsen\,x\cdot x-\int\!x\cdot \frac{1}{1-x^2}\,dx}\\\\\\ \mathtt{\int\!arcsen\,x\,dx=x\,arcsen\,x-\int\!\frac{x}{\sqrt{1-x^2}}\,dx}\\\\\\ \mathtt{\int\!arcsen\,x\,dx=x\,arcsen\,x-\int\!\frac{1}{-2}\cdot \dfrac{-2x}{\sqrt{1-x^2}}\,dx}\\\\\\ \mathtt{\int\!arcsen\,x\,dx=x\,arcsen\,x+\frac{1}{2}\int\!\dfrac{1}{\sqrt{1-x^2}}\cdot (-2x)\,dx}\\\\\\ \mathtt{\int\!arcsen\,x\,dx=x\,arcsen\,x+\frac{1}{2}\int\!\dfrac{1}{\sqrt{w}}\,dw\quad\quad\quad(w=1-x^2)}$

$\displaystyle\mathtt{\int\!arcsen\,x\,dx=x\,arcsen\,x+\frac{1}{2}\int\!\dfrac{1}{w^{1/2}}\,dw}\\\\\\ \mathtt{\int\!arcsen\,x\,dx=x\,arcsen\,x+\frac{1}{2}\int\!w^{-1/2}\,dw}\\\\\\ \mathtt{\int\!arcsen\,x\,dx=x\,arcsen\,x+\frac{1}{2}\cdot \frac{w^{-1/2+1}}{-\frac{1}{2}+1}+C}\\\\\\ \mathtt{\int\!arcsen\,x\,dx=x\,arcsen\,x+\frac{1}{2}\cdot \frac{w^{1/2}}{\frac{1}{2}}+C}\\\\\\ \mathtt{\int\!arcsen\,x\,dx=x\,arcsen\,x+\frac{1}{\diagup\!\!\!\! 2}\cdot \diagup\!\!\!\! 2\sqrt{w}+C}$


$\boxed{\displaystyle\mathtt{\int\!arcsen\,x\,dx=x\,arcsen\,x+\sqrt{1-x^2}+C}}$


Dúvidas? Comente.


Bons estudos! :-)