Cho hàm số \(f\left( x \right)\) liên tục trên \(\mathbb{R}\) và thỏa mãn \(f(x)+f(-x)=2\cos 2x,\,\forall x\in \mathbb{R}\). Khi đó \(\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}...

* Hướng dẫn giải

Với \(f(x)+f(-x)=2\cos 2x,\,\forall x\in \mathbb{R}\) \(\Rightarrow \int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\left( f(x)+f(-x) \right)\text{d}x}=\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{2\cos 2x\text{d}x}\Leftrightarrow \int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{f\left( x \right)\text{d}x}+\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{f\left( -x \right)\text{d}x}=\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{2\cos 2x\text{d}x}\) (*)

Tính \(I=\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{f\left( -x \right)\text{d}x}\)

Đặt \(t=-x\Rightarrow \text{d}t=-\text{d}x\Rightarrow \text{d}x=-\text{d}t\).

Đổi cận: \(x=\frac{\pi }{2}\Rightarrow t=-\frac{\pi }{2}\); \(x=-\frac{\pi }{2}\Rightarrow t=\frac{\pi }{2}\).

Khi đó \(I=-\int\limits_{\frac{\pi }{2}}^{-\frac{\pi }{2}}{f\left( t \right)\text{d}t}=\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{f\left( t \right)\text{d}t}=\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{f\left( x \right)\text{d}x}\).

Từ (*), ta được: \(2\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{2\cos 2x\text{d}x}=\left. \sin 2x \right|_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=0\) \(\Rightarrow \int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{f\left( x \right)\text{d}x}=0\).