Cho hàm số . Tích phân \(\int\limits_{0}^{\frac{\pi }{2}}{f\left( 2{{\sin }^{2}}x+3 \right)\sin 2x\text{d}x}\) bằng

* Hướng dẫn giải

Ta có

\(\begin{align} & \underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,\left( 2x-4 \right)=4;\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,\left( \frac{1}{4}{{x}^{3}}-{{x}^{2}}+x \right)=4;f\left( 4 \right)=4 \\ & \Rightarrow \underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 4 \right) \\ \end{align}\)

Nên hàm số đã cho liên tục tại \(x=4\)

Xét \(I=\int\limits_{0}^{\frac{\pi }{2}}{f\left( 2{{\sin }^{2}}x+3 \right)\sin 2x\text{d}x}\)

Đặt \(2{{\sin }^{2}}x+3=t\)\(\Rightarrow \)\(\sin 2x\text{d}x=\frac{1}{2}\text{d}t\)

Với \(x=0\)\(\Rightarrow t=3\)

\(x=\frac{\pi }{2}\)\(\Rightarrow t=5\)

\(\Rightarrow I=\int\limits_{3}^{5}{f\left( t \right)\frac{1}{2}\text{d}t}=\frac{1}{2}\int\limits_{3}^{5}{f\left( t \right)\text{d}t}=\frac{1}{2}\int\limits_{3}^{4}{\left( \frac{1}{4}{{t}^{3}}-{{t}^{2}}+t \right)\text{d}t}+\frac{1}{2}\int\limits_{4}^{5}{\left( 2t-4 \right)\text{d}t}=\frac{341}{96}\).