Cho hàm số \(f\left(x\right)\) liên tục trên \(\mathbb{R}\) thỏa mãn điều kiện \(f\left(x\right)+f\left(-x\right)=\sqrt{2+2\cos2x},\forall x\in\mathbb{R}\).
Tích phân \(\int_{-\frac{3\pi}{2}}^{\frac{3\pi}{2}}f\left(x\right)\text{d}x\) bằng
\(-6\). \(0\). \(6\). \(-2\). Hướng dẫn giải: Ta có \(I=\int_{-\frac{3\pi}{2}}^{\frac{3\pi}{2}}f\left(x\right)\text{d}x=\int_{-\frac{3\pi}{2}}^0f\left(x\right)\text{d}x+\int_0^{\frac{3\pi}{2}}f\left(x\right)\text{d}x\) (1)Đặt \(t=-x\) thì \(x=-t.\text{d}x=-\text{d}t,f\left(x\right)=f\left(-t\right)\). Đổi cận: Khi \(x=-\dfrac{3\pi}{2}\)thì \(t=\dfrac{3\pi}{2}\), khi \(x=0\) thì \(t=0\) nên \(\int_{-\frac{3\pi}{2}}^0f\left(x\right)\text{d}x=\int_{\frac{3\pi}{2}}^0f\left(-t\right)\left(-\text{d}t\right)=\int_0^{\frac{3\pi}{2}}f\left(-t\right)\text{d}t\)
Mà \(\int_0^{\frac{3\pi}{2}}f\left(-t\right)\text{d}t=\int_0^{\frac{3\pi}{2}}f\left(-x\right)\text{d}x\) (tích phân xác định không phụ thuộc kí hiệu biến lấy tích phân) nên \(\int_{-\frac{3\pi}{2}}^0f\left(x\right)\text{d}x=\int_0^{\frac{3\pi}{2}}f\left(-x\right)\text{d}x\) (2). Từ (1) và (2) suy ra
\(I=\int_0^{\frac{3\pi}{2}}f\left(x\right)\text{d}x+\int_0^{\frac{3\pi}{2}}f\left(-x\right)\text{d}x=\int_0^{\frac{3\pi}{2}}\left(f\left(x\right)+f\left(-x\right)\right)\text{d}x=\int_0^{\frac{3\pi}{2}}\sqrt{2+2\cos2x}\text{d}x\)
\(=\int_0^{\frac{3\pi}{2}}\sqrt{4\cos^2x}\text{d}x=\int_0^{\frac{3\pi}{2}}2\left|\cos x\right|\text{d}x=\int_0^{\frac{\pi}{2}}2\cos x\text{d}x-\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}2\cos x\text{d}x\)
\(=2\sin x|_0^{\frac{\pi}{2}}-2\sin x|_{\frac{\pi}{2}}^{\frac{3\pi}{2}}=2\left(1-0\right)-2\left(-1-1\right)=6\).