Question

1)\(\int_1^e\left(\frac{lnx}{x}\right)^2dx\)

2)\(\int_0^{\frac{\pi}{4}}\frac{x}{1+cos2x}dx\)

3)\(\int_0^{\frac{\pi}{4}}\frac{ln\left(cosx\right)}{cos^2x}dx\)

Answer 1

Câu 1)

Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2\ln x}{x}\\ v=\frac{-1}{x}\end{matrix}\right.\)

\(\int \left ( \frac{\ln}{x} \right )^2dx=\frac{-\ln^2x}{x}+2\int \frac{\ln x}{x^2}dx\)

Đặt \(\left\{\begin{matrix} t=\ln x\\ dk=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dt=\frac{1}{x}dx\\ k=-\frac{1}{x}\end{matrix}\right.\Rightarrow \int \frac{\ln x}{x^2}dx=-\frac{\ln x}{x}+\int \frac{1}{x^2}dx=\frac{-\ln x}{x}-\frac{1}{x}\)

\(\Rightarrow I=\left.\begin{matrix} e\\ 1\end{matrix}\right|\left(\frac{-\ln^2 x}{x}-\frac{2\ln x}{x}-\frac{2}{x}\right)=2-\frac{5}{e}\)

Answer 2

Câu 2)

\(I=\int ^{\frac{\pi}{4}}_{0}\frac{x}{1+\cos 2x}dx=\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{x}{\cos^2x}dx\)

Đặt \(\left\{\begin{matrix} u=x\\ dv=\frac{dx}{\cos^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\tan x\end{matrix}\right.\Rightarrow I=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{x\tan x}{2}-\frac{1}{2}\int^{\frac{\pi}{4}}_{0} \tan xdx\)

\(=\frac{\pi}{8}+\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{d(\cos x)}{\cos x}=\frac{\pi}{8}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{\ln |\cos x|}{2}=\frac{\pi}{8}+\frac{\ln\frac{\sqrt{2}}{2}}{2}\)

Answer 3

Câu 3)

Đặt \(\left\{\begin{matrix} u=\ln (\cos x)\\ dv=\frac{dx}{\cos^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{-\sin x}{\cos x}dx=-\tan xdx\\ v=\tan x\end{matrix}\right.\)

\(\Rightarrow I=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\tan x\ln (\cos x)+\int ^{\frac{\pi}{4}}_{0}\tan^2xdx=\ln \frac{\sqrt{2}}{2}+\int ^{\frac{\pi}{4}}_{0}(\frac{1}{\cos^2x}-1)dx\)

\(=\ln\frac{\sqrt{2}}{2}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|(\tan x-x)=\ln \frac{\sqrt{2}}{2}-\frac{\pi}{4}+1\)