Giá trị của tích phân $\Large I=\int\limits_{0}^{\dfrac{\pi }{3}}{\dfr

Giá trị của tích phân $\Large I=\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\cos x}{\sqrt{2+\cos 2x}}dx}$ là

• A.

  A. $\Large \dfrac{\pi }{4\sqrt{2}}$

• B.

  B. $\Large \dfrac{\pi }{2\sqrt{2}}$

• C.

  C. $\Large \dfrac{4\pi }{\sqrt{2}}$

• D.

  D. $\Large \dfrac{-\pi }{\sqrt{2}}$

Đặt $\Large t=\sin x\Rightarrow dt=\cos xdx$ . Đổi cận $\Large \left\{ \begin{align}  & x=0\Rightarrow t=0 \\  & x=\dfrac{\pi }{3}\Rightarrow t=\dfrac{\sqrt{3}}{2} \\ \end{align} \right.$

Vậy $\Large I=\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\cos x}{\sqrt{2+\cos 2x}}dx=\int\limits_{0}^{\dfrac{\sqrt{3}}{2}}{\dfrac{dt}{\sqrt{3-2{{t}^{2}}}}=\dfrac{1}{\sqrt{2}}}}\int\limits_{0}^{\dfrac{\sqrt{3}}{2}}{\dfrac{dt}{\sqrt{\dfrac{3}{2}-{{t}^{2}}}}}$

Đặt $\Large t=\sqrt{\dfrac{3}{2}}\cos u\Rightarrow dt=-\sqrt{\dfrac{3}{2}}\sin udu$ . Đổi cận $\Large \left\{ \begin{align}  & t=0\Rightarrow u=\dfrac{\pi }{2} \\  & t=\dfrac{\sqrt{3}}{2}\Rightarrow u=\dfrac{\pi }{4} \\ \end{align} \right.$

Suy ra $\Large I = \dfrac{1}{\sqrt{2}}\int\limits_{0}^{\dfrac{\sqrt{3}}{2}}{\dfrac{dt}{\sqrt{\dfrac{3}{2}-{{t}^{2}}}}=\dfrac{1}{\sqrt{2}}}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\dfrac{\sqrt{3}}{2}\sin udu}{\sqrt{\dfrac{3}{2}(1-{{\cos }^{2}}u)}}=}$ $\Large \dfrac{1}{\sqrt{2}}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{du=\dfrac{1}{\sqrt{2}}}u\left| \begin{align}  & \dfrac{\pi }{2} \\  & \dfrac{\pi }{4} \\ \end{align} \right.=\dfrac{\pi }{4\sqrt{2}}$