|
|
Ta biết rằng
$(\arcsin x)' = \dfrac{1}{\sqrt{1-x^2}}$
Suy ra
$\arcsin x=\arcsin x+\dfrac{x}{\sqrt{1-x^2}}-\dfrac{x}{\sqrt{1-x^2}}=(x \arcsin x)'+\left (\sqrt{1-x^2} \right )'$
Vậy
$\int\limits_{0}^{1/2}\arcsin xdx = \left[ {x \arcsin x+\sqrt{1-x^2}} \right]_{0}^{1/2}=\boxed{\dfrac{\sqrt 3-2}{2}+\dfrac{\pi }{12}}$
|