Trong các khẳng định sau, khẳng định nào sai ?
\(\left(\dfrac{1}{\cos x}\right)'=\dfrac{\sin x}{\cos^2x}\).\(\int\limits^{\frac{\pi}{3}}_0\dfrac{x\sin x}{\cos^2x}\text{d}x=\dfrac{x}{\cos x}\bigg|^{\frac{\pi}{3}}_0-\int\limits^{\frac{\pi}{3}}_0\dfrac{1}{\cos x}\text{d}x\).\(\int\limits^{\frac{\pi}{3}}_0\dfrac{1}{\cos x}\text{d}x=\int\limits^{\frac{\pi}{3}}_0\dfrac{\text{d}\left(\sin x\right)}{\sin^2x}=\dfrac{1}{2}\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\bigg|^{\frac{\pi}{3}}_0\).\(\int\limits^{\frac{\pi}{3}}_0\dfrac{x\sin x}{\cos^2x}\text{d}x=\dfrac{2\pi}{3}-\ln\left(2-\sqrt{3}\right)\).Hướng dẫn giải:Ta có:
*) \(\left(\dfrac{1}{\cos x}\right)'=\dfrac{-\left(\cos x\right)'}{\cos^2x}=\dfrac{\sin x}{\cos^2x}\), sử dụng công thức này để tính tích phân từng phần trong tính tích phân các câu còn lại.
*) Đặt \(\begin{cases}u=x\\v'=\dfrac{\sin x}{\cos^2x}\end{cases}\) Theo công thức trên suy ra \(\begin{cases}u'=1\\v=\dfrac{1}{\cos x}\end{cases}\)
Vậy \(\int\limits^{\frac{\pi}{3}}_0\dfrac{x\sin x}{\cos^2x}dx=\dfrac{x}{\cos x}|^{\frac{\pi}{3}}_0-\int\limits^{\frac{\pi}{3}}_0\dfrac{1}{\cos x}dx\)
Ta lại tính:
\(\int\limits^{\frac{\pi}{3}}_0\dfrac{1}{\cos x}\text{d}x=\int\limits^{\frac{\pi}{3}}_0\dfrac{\cos x}{\cos^2x}\text{d}x=\int\limits^{\frac{\pi}{3}}_0\dfrac{\text{d}\left(\sin x\right)}{1-\sin^2x}\)
\(=\int\limits^{\frac{\pi}{3}}_0\dfrac{1}{2}\left[\dfrac{1}{1-\sin x}+\dfrac{1}{1+\cos x}\right]\text{d}\left(\sin x\right)\)
\(=\) \(\dfrac{1}{2}\left[-\int\limits^{\frac{\pi}{3}}_0\dfrac{\text{d}\left(1-\sin x\right)}{1-\sin x}+\int\limits^{\frac{\pi}{3}}_0\dfrac{\text{d}\left(1+\sin x\right)}{1+\sin x}\right]\)
\(=\dfrac{1}{2}\left[-\ln\left|1-\sin x\right|+\ln\left|1+\sin x\right|\right]|^{\frac{\pi}{3}}_0\)
\(=\dfrac{1}{2}\ln\dfrac{1+\sin x}{1-\sin x}|^{\frac{\pi}{3}}_0\)
Vậy: \(\int\limits^{\frac{\pi}{3}}_0\dfrac{x\sin x}{\cos^2x}\text{d}x=\dfrac{x}{\cos x}|^{\frac{\pi}{3}}_0-\dfrac{1}{2}\ln\dfrac{1+\sin x}{1-\sin x}|^{\frac{\pi}{3}}_0\)
\(=\dfrac{\frac{\pi}{3}}{\frac{1}{2}}-\dfrac{1}{2}\ln\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\)
\(=\dfrac{2\pi}{3}-\dfrac{1}{2}\ln3\)
Vậy \(\int\limits^{\frac{\pi}{3}}_0\dfrac{x\sin x}{\cos^2x}\text{d}x=\dfrac{2\pi}{3}-\ln\left(2-\sqrt{3}\right)\) là khẳng định sai.