Đáp án:
\[I = \pi - 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
I = \int\limits_0^{\frac{\pi }{2}} {\left( {2x + 1} \right)\cos xdx} \\
\Rightarrow \left\{ \begin{array}{l}
u = 2x + 1\\
dv = \cos xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 2\\
v = \sin x
\end{array} \right.\\
\Rightarrow I = \mathop {\left. {\left( {2x + 1} \right)\sin x} \right|}\nolimits_0^{\frac{\pi }{2}} - \int\limits_0^{\frac{\pi }{2}} {2\sin xdx} \\
= \left( {2.\frac{\pi }{2} + 1} \right).\sin \frac{\pi }{2} - \left( {2.0 + 1} \right).\sin 0 - \mathop {\left. {\left( { - 2\cos x} \right)} \right|}\nolimits_0^{\frac{\pi }{2}} \\
= \left( {\pi + 1} \right) + 2\left( {\cos \frac{\pi }{2} - \cos 0} \right)\\
= \pi + 1 + 2.\left( { - 1} \right) = \pi - 1
\end{array}\)
