Đáp án:
\(\dfrac{1}{{29}}\)
Giải thích các bước giải:
\(\begin{array}{l}
I = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^{20}}x{{\cos }^8}xdx} = \int\limits_0^{\frac{\pi }{2}} {{{\left( {{{\sin }^9}x} \right)}^{\frac{{20}}{9}}}{{\cos }^8}xdx} \\
Dat\,\,t = {\sin ^9}x\\
\Rightarrow dt = 9{\cos ^8}xdx \Leftrightarrow {\cos ^8}xdx = \frac{1}{9}dt\\
Doi\,\,can:\,\,\left\{ \begin{array}{l}
x = 0 \Rightarrow t = 0\\
x = \frac{\pi }{2} \Rightarrow t = 1
\end{array} \right.\\
\Rightarrow I = \frac{1}{9}\int\limits_0^1 {{t^{\frac{{20}}{9}}}dt} = \frac{1}{9}\left. {\frac{{{t^{\frac{{29}}{9}}}}}{{\frac{{29}}{9}}}} \right|_0^1 = \frac{1}{{29}}\left. {{t^{\frac{{29}}{9}}}} \right|_0^1 = \frac{1}{{29}}
\end{array}\)
