Cho hàm số \(f(x)\) có đạo hàm liên tục trên \(\left[ {0;\frac{\pi }{2}} \right],\) thỏa mãn \(\int\limits_0^{\frac{\pi }{2}}

* Hướng dẫn giải

Xét \(\int\limits_0^{\frac{\pi }{2}} {f'\left( x \right){{\cos }^2}x{\rm{d}}x}  = 10\), đặt \(\left\{ \begin{array}{l}
u = {\cos ^2}x\\
{\rm{d}}v = f'\left( x \right){\cos ^2}x{\rm{d}}x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\rm{d}}u =  - \sin 2x{\rm{d}}x\\
v = f\left( x \right)
\end{array} \right..\)

Khi đó \(10 = \int\limits_0^{\frac{\pi }{2}} {f'\left( x \right){{\cos }^2}x{\rm{d}}x}  = {\cos ^2}xf\left( x \right)\left| {\begin{array}{*{20}{c}}
{\frac{\pi }{2}}\\
0
\end{array}} \right. + \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\sin 2x{\rm{d}}x} {\rm{ }}\)

\( \Leftrightarrow 10 =  - f\left( 0 \right) + \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\sin 2x{\rm{d}}x}  \to \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\sin 2x{\rm{d}}x}  = 10 + f\left( 0 \right) = 13.\)