Trong các khẳng định sau, khẳng định nào sai ?
\(\left(\tan x-x\right)'=\tan^2x\). \(\int\limits^{\frac{\pi}{4}}_0x\tan^2xdx=x\left(\tan x-x\right)|^{\frac{\pi}{4}}_0-\int\limits^{\frac{\pi}{4}}_0\left(\tan x-x\right)dx\). \(\int\limits^{\frac{\pi}{4}}_0x\tan^2xdx=\dfrac{\pi}{4}\left(1-\dfrac{\pi}{4}\right)+\int\limits^{\frac{\pi}{4}}_0\dfrac{d\cos x}{\cos x}+\int\limits^{\frac{\pi}{4}}_0xdx\). \(\int\limits^{\frac{\pi}{4}}_0x\tan^2dx=\dfrac{\pi}{4}+\dfrac{\pi^2}{32}-\dfrac{1}{2}\ln2\). Hướng dẫn giải:Ta kiểm tra:
*) \(\left(\tan x-x\right)'=\dfrac{1}{\cos^2x}-1=\dfrac{1-\cos^2x}{\cos^2x}=\dfrac{\sin^2x}{\cos^2x}=\tan^2x\)
*) Đặt \(\begin{cases}u=x\\v'=\tan^2x\end{cases}\) suy ra \(\begin{cases}u'=1\\v=\int\tan^2x\text{d}x=\tan x-x\end{cases}\)
(chú ý dựa vào ý trên \(\left(\tan x-x\right)'=\tan^2x\))
Vậy \(\int\limits^{\frac{\pi}{4}}_0x\tan^2xdx=x\left(\tan x-x\right)|^{\frac{\pi}{4}}_0-\int\limits^{\frac{\pi}{4}}_0\left(\tan x-x\right)dx\)
\(=\dfrac{\pi}{4}\left(1-\dfrac{\pi}{4}\right)-\int\limits^{\frac{\pi}{4}}_0\dfrac{-\text{d}\left(\cos x\right)}{\cos x}+\int\limits^{\frac{\pi}{4}}_0x\text{d}x\)
\(=\dfrac{\pi}{4}\left(1-\dfrac{\pi}{4}\right)+\ln\cos x|^{\frac{\pi}{4}}_0+\dfrac{x^2}{2}|^{\frac{\pi}{4}}_0\)
\(=\dfrac{\pi}{4}-\dfrac{\pi^2}{32}-\dfrac{1}{2}\ln2\)
Vậy \(\int\limits^{\frac{\pi}{4}}_0x\tan^2dx=\dfrac{\pi}{4}+\dfrac{\pi^2}{32}-\dfrac{1}{2}\ln2\) là khẳng định sai.