Cho hs \(f\left( x \right) = \left\{ \begin{array}{l}{x^2} – 1{\rm{ }}\\{x^2} – 2x + 3\end{array} \right.

Xét \(I=\int\limits_{0}^{\frac{\pi }{2}}{f\left( 2\sin x+1 \right)\text{cosxdx}}.\)

Đặt \(t=2\operatorname{s}\text{inx+1}\) ta có \(dt=2\cos xdx.\)

Đổi cận: \(\left\{ \begin{align}

& x=0\Rightarrow t=1 \\

& x=\frac{\pi }{2}\Rightarrow t=3 \\

\end{align} \right..\) Khi đó ta có:

\(I=\frac{1}{2}\int\limits_{1}^{3}{f\left( t \right)dt=\frac{1}{2}}\int\limits_{1}^{3}{f\left( x \right)dx}\)

\(=\frac{1}{2}\left( \int\limits_{1}^{2}{f\left( x \right)dx+\int\limits_{2}^{3}{f\left( x \right)dx}} \right)\)

\(=\frac{1}{2}\left( \int\limits_{1}^{2}{\left( {{x}^{2}}-2x+3 \right)dx+\int\limits_{2}^{3}{\left( {{x}^{2}}-1 \right)dx}} \right)\)

\(=\frac{1}{2}\left( \frac{7}{3}+\frac{16}{3} \right)=\frac{23}{6}\)