Tích phân \(I = \int\limits_0^{\frac{\pi }{4}} {\f...

B

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

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