Tích phân \(I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\left( {{x^3} + 2x} \right)\cos x + x{{\cos }^2}x}}{{\cos x}}dx} \) có giá trị là:

* Hướng dẫn giải

Ta có:

\(\begin{array}{l} I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\left( {{x^3} + 2x} \right)\cos x + x{{\cos }^2}x}}{{\cos x}}dx} \\ = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\left( {{x^3} + 2x} \right)dx} + \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {x\cos xdx} \\ = \left. {\left( {\frac{1}{4}{x^4} + {x^2}} \right)} \right|_{\frac{\pi }{6}}^{\frac{\pi }{2}} + \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {x\cos xdx} \end{array}\)

Xét \({I_1} = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {x\cos xdx} \).

Đặt \(\left\{ \begin{array}{l} u = x\\ dv = \cos xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = dx\\ v = \sin x \end{array} \right.\).

\(\Rightarrow {I_1} = \left. {\left( {x\sin x} \right)} \right|_{\frac{\pi }{6}}^{\frac{\pi }{2}} - \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sin xdx} = \frac{\pi }{4} - \frac{{\sqrt 3 }}{2}\).

\( \Rightarrow I = \left. {\left( {\frac{1}{4}{x^4} + {x^2}} \right)} \right|_{\frac{\pi }{6}}^{\frac{\pi }{2}} + {I_1} = \frac{{5{\pi ^4}}}{{324}} + \frac{{2{\pi ^2}}}{9} + \frac{\pi }{4} - \frac{{\sqrt 3 }}{2}\).