Tích phân \(I = \int\limits_0^{\frac{\pi }{4}} {\frac{{2x - \sin x}}{{2 - 2\cos x}}dx} \) có giá trị là:

* Hướng dẫn giải

Ta biến đổi:

\(I = \int\limits_{\frac{\pi }{3}}^{\frac{\pi }{4}} {\frac{{2x - \sin x}}{{2 - 2\cos x}}dx} = \int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\frac{x}{{1 - \cos x}}dx} - \frac{1}{2}\int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\frac{{\sin x}}{{1 - \cos x}}dx} \)

Xét \({I_1} = \int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\frac{x}{{1 - \cos x}}dx} = \frac{1}{2}\int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\frac{x}{{{{\sin }^2}\frac{x}{2}}}dx} \).

Đặt \(\left\{ \begin{array}{l} u = x\\ dv = \frac{1}{{{{\sin }^2}\frac{x}{2}}}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = dx\\ v = - 2\cot \frac{x}{2} \end{array} \right.\).

\( \Rightarrow {I_1} = \frac{1}{2}\left[ {\left. {\left( { - 2x.\cot \frac{x}{2}} \right)} \right|_{\frac{\pi }{3}}^{\frac{\pi }{2}} + 2\int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\cot \frac{x}{2}dx} } \right] \\= \frac{1}{2}\left[ { - \pi + \frac{{2\pi \sqrt 3 }}{3} + 4\ln \sqrt 2 } \right]\).

Xét \({I_2} = \frac{1}{2}\int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\frac{{\sin x}}{{1 - \cos x}}dx} \).

Đặt \(t = 1 - \cos x \Rightarrow dt = \sin xdx\).

Đổi cận \(\left\{ \begin{array}{l} x = \frac{\pi }{3} \Rightarrow t = \frac{1}{2}\\ x = \frac{\pi }{2} \Rightarrow t = 1 \end{array} \right.\).

\(\Rightarrow {I_2} = \frac{1}{2}\int\limits_{\frac{1}{2}}^1 {\frac{1}{t}dt = \frac{1}{2}\left. {\left( {\ln \left| t \right|} \right)} \right|} _{\frac{1}{2}}^1 = \frac{1}{2}\ln 2\).

\(I = {I_1} - {I_2} = \frac{1}{2}\left( { - \pi + \frac{{2\pi \sqrt 3 }}{3} + 4\ln \sqrt 2 - \ln 2} \right)\).