Biết \(\int\limits_0^\pi {\frac{{x{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} = \frac{{{\pi ^a}}}{b}\) với \(a,b \in {Z^ + }.\) Tính \(P = 2a + b.\)
Suy nghĩ trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo saiGọi \(I = \int\limits_0^\pi {\frac{{x{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} \)
Đặt \(t = \pi - x \to {\rm{d}}t = - {\rm{d}}x.\) Đổi cận \(\left\{ \begin{array}{l}
x = 0 \to t = \pi \\
x = \pi \to t = 0
\end{array} \right..\)
Khi đó \(I = - \int\limits_\pi ^0 {\frac{{\left( {\pi - t} \right){{\sin }^{2018}}\left( {\pi - t} \right)}}{{{{\sin }^{2018}}\left( {\pi - t} \right) + {{\cos }^{2018}}\left( {\pi - t} \right)}}{\rm{d}}t} = \int\limits_0^\pi {\frac{{\left( {\pi - t} \right){{\sin }^{2018}}t}}{{{{\sin }^{2018}}t + {{\cos }^{2018}}t}}{\rm{d}}t} = \int\limits_0^\pi {\frac{{\left( {\pi - x} \right){{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} .\)
Suy ra \(2I = \int\limits_0^\pi {\frac{{x{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} + \int\limits_0^\pi {\frac{{\left( {\pi - x} \right){{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} = \int\limits_0^\pi {\frac{{\pi {{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} \)
\( \to I = \frac{\pi }{2}\int\limits_0^\pi {\frac{{{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} = \frac{\pi }{2}\left[ {\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} + \int\limits_{\frac{\pi }{2}}^\pi {\frac{{{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} } \right].\)
Đặt \(x = u + \frac{\pi }{2}\) ta suy ra \(\int\limits_{\frac{\pi }{2}}^\pi {\frac{{{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{2018}}u}}{{{{\sin }^{2018}}u + {{\cos }^{2018}}u}}{\rm{d}}u} = \int\limits_{\frac{\pi }{2}}^\pi {\frac{{{{\cos }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} .\)
Vậy \(I = \frac{\pi }{2}\int\limits_0^{\frac{\pi }{2}} {{\rm{d}}x} = \frac{{{\pi ^2}}}{4} \to \left\{ \begin{array}{l}
a = 2\\
b = 4
\end{array} \right. \to P = 8.\)
