Cho \(I=\int\limits^6_{3\sqrt{2}}\frac{\text{d}x}{x\sqrt{x^2-9}}\) , đặt \(x=\frac{3}{\cos t}\) . Khẳng định nào sau đây sai?
\(\text{d}x=\frac{3\sin t}{\cos^2t}\text{d}t\). \(\frac{\text{d}x}{x\sqrt{x^2-9}}=\frac{\sin t\text{d}t}{3\cos t\tan t}\). \(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\sin t\text{d}t}{3\cos t\tan t}\). \(I=\frac{\pi}{36}\). Hướng dẫn giải:Đặt \(x=\frac{3}{\cos t}\) suy ra \(\text{d}x=\dfrac{-3.\left(-\sin t\right)}{\cos^2t}\text{d}t=\dfrac{3\sin t}{\cos^2t}\text{d}t\)
Đổi cận: \(x|^6_{3\sqrt{2}}\Rightarrow t|^{\frac{\pi}{3}}_{\frac{\pi}{4}}\) (vì \(\frac{3}{\cos t}=3\sqrt{2}\Rightarrow t=\frac{\pi}{4};\frac{3}{\cos t}=6\Rightarrow t=\frac{\pi}{3}\))
Vậy:
\(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\frac{3\sin t}{\cos^2t}\text{d}t}{\frac{3}{\cos t}\sqrt{\frac{9}{\cos^2t}-9}}\)
\(=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\sin t}{3\cos t.\tan t}\text{d}t\)
\(=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{1}{3}\text{d}t\)
\(=\frac{1}{3}t|^{\frac{\pi}{3}}_{\frac{\pi}{4}}\)
\(=\frac{1}{3}\left(\frac{\pi}{3}-\frac{\pi}{4}\right)=\frac{\pi}{36}\)
Chú ý: \(\frac{\text{d}x}{x\sqrt{x^2-9}}=\frac{\sin t\text{d}t}{3\cos t.\left|\tan t\right|}\) và \(\frac{\text{d}x}{x\sqrt{x^2-9}}=\frac{\sin t\text{d}t}{3\cos t\tan t}\) trong trường hợp \(x\in\left[\frac{\pi}{4};\frac{\pi}{3}\right]\).