Cho hàm số \(f(x)\) có đạo hàm liên tục trên \([0;\pi ]\).

* Hướng dẫn giải

\(\begin{array}{l}
f'\left( x \right) + \sin \,xf\left( x \right) = \cos x{e^{\cos x}}\,\,\,\forall x \in \left[ {0;\pi } \right]\\
 \Leftrightarrow f'\left( x \right){e^{ - \cos x}} + \sin \,xf\left( x \right){e^{ - \cos x}} = \cos x\\
 \Leftrightarrow \left[ {f\left( x \right){e^{ - \cos x}}} \right]' = \cos x\\
 \Leftrightarrow \int\limits_0^x {\left[ {f\left( x \right){e^{ - \cos x}}} \right]dx = \int\limits_0^x {\cos xdx} } \\
 \Leftrightarrow f\left( x \right){e^{ - \cos x}}\left| \begin{array}{l}
^x\\
_0
\end{array} \right. = \sin \,x\left| \begin{array}{l}
^x\\
_0
\end{array} \right.\\
 \Leftrightarrow f\left( x \right){e^{ - \cos x}} - f\left( 0 \right).{e^{ - 1}} = \sin \,x\\
 \Leftrightarrow f\left( x \right){e^{ - \cos x}} - 2e.{e^{ - 1}} = \sin \,x\\
 \Leftrightarrow f\left( x \right){e^{ - \cos x}} = \sin \,x + 2\\
 \Leftrightarrow f\left( x \right) = \left( {\sin \,x + 2} \right){e^{\cos x}}
\end{array}\)

Khi đó ta có \(I = \int\limits_0^\pi  {f\left( x \right)dx = \int\limits_0^\pi  {\left( {\sin \,x + 2} \right){e^{\cos x}}dx \approx 10,31} } \)