Đáp án:
\[I = \frac{2}{3}.\left( {27 - 16\sqrt 2 } \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
t = 8 + \cos x \Rightarrow \left\{ \begin{array}{l}
dt = \left( {8 + \cos x} \right)'dx = - \sin xdx\\
x = 0 \Rightarrow t = 9\\
x = \frac{\pi }{2} \Rightarrow t = 8
\end{array} \right.\\
\Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\sin x\sqrt {8 + \cos x} dx} \\
= - \int\limits_0^{\frac{\pi }{2}} {\sqrt {8 + \cos x} \left( { - \sin xdx} \right)} \\
= - \int\limits_9^8 {\sqrt t dt} \\
= \int\limits_8^9 {\sqrt t dt} \\
= \int\limits_8^9 {{t^{\frac{1}{2}}}dt} \\
= \mathop {\left. {\frac{{{t^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}}} \right|}\nolimits_8^9 \\
= \frac{2}{3}.\left( {{9^{\frac{3}{2}}} - {8^{\frac{3}{2}}}} \right) = \frac{2}{3}.\left( {27 - 16\sqrt 2 } \right)
\end{array}\)
