Tích phân $\Large I=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{2x-\sin x

Tích phân $\Large I=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{2x-\sin x}{2-2\cos x}dx}$ có giá trị là 

• A.

  $\Large I=\dfrac{1}{2}\left( -\pi +\dfrac{2\pi \sqrt{3}}{3}+4\ln \sqrt{2}+\ln 2 \right)$

• B.

  $\Large I=\dfrac{1}{2}\left( -\pi +\dfrac{2\pi \sqrt{3}}{3}+2\ln \sqrt{2}-\ln 2 \right)$

• C.

  $\Large I=\dfrac{1}{2}\left( -\pi +\dfrac{2\pi \sqrt{3}}{3}+4\ln \sqrt{2}-\ln 2 \right)$

• D.

  $\Large I=\dfrac{1}{2}\left( -\pi +\dfrac{2\pi \sqrt{3}}{3}+2\ln \sqrt{2}+\ln 2 \right)$

Tích phân $\Large I=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{2x-\sin x}{2-2\cos x}dx}$ có giá trị là:

Ta biến đổi: $\Large I=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{2x-\sin x}{2-2\cos x}dx=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{x}{1-\cos x}dx-\dfrac{1}{2}\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1-\cos x}dx}}}$

Xét $\Large {{I}_{1}}=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{x}{1-\cos x}dx=\dfrac{1}{2}\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{x}{{{\sin }^{2}}\dfrac{x}{2}}dx}}$

Đặt $\Large \left\{ \begin{align}  & u=x \\  & dv=\dfrac{1}{{{\sin }^{2}}\dfrac{x}{2}}dx \\ \end{align} \right.$ $\Large \Rightarrow \left\{ \begin{align}  & du=dx \\  & v=-2\cot \dfrac{x}{2} \\ \end{align} \right.$

$\Large \Rightarrow {{I}_{1}}=\dfrac{1}{2}\left[ \left( -2x.\cot \dfrac{x}{2} \right)\left| \begin{align}  & \dfrac{\pi }{2} \\  & \dfrac{\pi }{3} \\ \end{align} \right.+2\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\cot \dfrac{x}{2}dx} \right]$ $\Large =\dfrac{1}{2}\left[ -\pi +\dfrac{2\pi \sqrt{3}}{3}+4\ln \sqrt{2} \right]$

Xét $\Large {{I}_{2}}=\dfrac{1}{2}\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1-\cos x}dx}$

Đặt $\Large t=1-\cos x\Rightarrow dt=\sin xdx$

Đổi cận: $\Large \left\{ \begin{align}  & x=\dfrac{\pi }{3}\Rightarrow t=\dfrac{1}{2} \\  & x=\dfrac{\pi }{2}\Rightarrow t=1 \\ \end{align} \right.$

$\Large \Rightarrow {{I}_{2}}=\dfrac{1}{2}\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{1}{t}dt=\dfrac{1}{2}\left( \ln \left| t \right| \right)\left| \begin{align}  & 1 \\  & \dfrac{1}{2} \\ \end{align} \right.=\dfrac{1}{2}\ln 2}$

$\Large I={{I}_{1}}-{{I}_{2}}=\dfrac{1}{2}\left( -\pi +\dfrac{2\pi \sqrt{3}}{3}+4\ln \sqrt{2}-\ln 2 \right)$

Chọn C