Cho $\Large I=\int\limits_{0}^{\dfrac{1}{2}}{\sqrt{1-2x\sqrt{1-{{x}^{2

Cho $\Large I=\int\limits_{0}^{\dfrac{1}{2}}{\sqrt{1-2x\sqrt{1-{{x}^{2}}}}}dx=a\pi +b$ với $\Large a,b\in R$ . Giá trị $\Large a+b$ gần nhất với 

• A.

  $\Large \dfrac{1}{10}$

• B.

  $\Large 1$

• C.

  $\Large \dfrac{1}{5}$

• D.

  $\Large \sqrt{2}$

Đặt $\Large x=\sin t,t\in \left[ -\dfrac{\pi }{2};\dfrac{\pi }{2} \right]\Rightarrow d=\cos tdt$ .

Đổi cận : $\Large x=0\Rightarrow t= 0; x=\dfrac{1}{2}\Rightarrow t=\dfrac{\pi }{6}$

Khi đó: $\Large I$ được viết lại là 

$\Large I=\int\limits_{0}^{\dfrac{\pi }{6}}{\sqrt{1-2\sin t\cos t}.\cos tdt=\int\limits_{0}^{\dfrac{\pi }{6}}{\sqrt{{{(\cos t-\sin t)}^{2}}}.\cos tdt=\int\limits_{0}^{\dfrac{\pi }{6}}{(\cos t-\sin t)\cos tdt}}}$

$\Large \Leftrightarrow I= -\int\limits_{0}^{\dfrac{\pi }{6}}{\sin t\cos tdt+\int\limits_{0}^{\dfrac{\pi }{6}}{{{\cos }^{2}}tdt}}=\dfrac{-1}{4}\int\limits_{0}^{\dfrac{\pi }{6}}{\sin 2td(2t)+\dfrac{1}{4}}\int\limits_{0}^{\dfrac{\pi }{6}}{(\cos 2t+1)d(2t)}$

$\Large \Leftrightarrow I=\dfrac{\cos 2t}{4}\left| \begin{align}  & \dfrac{\pi }{6} \\  & 0 \\ \end{align} \right.$ $\Large +\dfrac{\sin 2t+2t}{4}\left| \begin{align}  & \dfrac{\pi }{6} \\  & 0 \\ \end{align} \right.=$ $\Large \dfrac{\pi }{12}+\dfrac{\sqrt{3}-1}{8}=a\pi +b$

Suy ra $\Large a+b= \dfrac{1 }{12}+\dfrac{\sqrt{3}-1}{8}\approx 0,175$

Đáp án C