Tích phân \(I = \int\limits_0^{\frac{\pi }{4}} {\frac{x}{{1 + \cos x}}dx} \) có giá trị là:

* Hướng dẫn giải

Ta biến đổi: \(I = \int\limits_0^{\frac{\pi }{4}} {\frac{x}{{1 + \cos x}}dx} = \frac{1}{2}I\int\limits_0^{\frac{\pi }{4}} {\frac{x}{{{{\cos }^2}\frac{x}{2}}}dx} \).

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

\(\Rightarrow I = \frac{1}{2}\left[ {\left. {\left( {2x\tan \frac{x}{2}} \right)} \right|_0^{\frac{\pi }{4}} - 2\int\limits_0^{\frac{\pi }{4}} {\tan \frac{x}{2}dx} } \right] \\= \frac{1}{2}\left( {\frac{\pi }{2}\tan \frac{\pi }{8} - 2\int\limits_0^{\frac{\pi }{4}} {\frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}dx} } \right) \\= \frac{\pi }{2}\tan \frac{\pi }{8} + 4\int\limits_1^{\cos \frac{\pi }{8}} {\frac{1}{t}dt} \\= \frac{\pi }{4}\tan \frac{\pi }{8} + 2\ln \left( {\cos \frac{\pi }{8}} \right)\)