Tính tích phân }{2} - a} {{\sin }{2} > a > 0 \)

* Hướng dẫn giải

Ta có:

\(\int\limits_a^{\dfrac{\pi }{2} - a} {{{\cos }^2}x\,dx\,} \)

\(= \dfrac{1}{2}\int\limits_a^{\dfrac{\pi }{2} - a} {\dfrac{{\cos 2x + 1}}{2}} \,d\left( {2x} \right) \)

\(= \dfrac{1}{4}\left( {\sin 2x + 2x} \right)\left| \begin{array}{l}^{\dfrac{\pi }{2} - a}\\_a\end{array} \right. \)

\(= \dfrac{1}{4}\left( {\sin \left( {\pi  - 2a} \right) + \pi  - 2a - \sin 2a - 2a} \right)\)

\( = \dfrac{1}{4}\left( {\sin \left( {\pi  - 2a} \right) - \sin 2a + \pi  - 4a} \right)\)