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AH
Akai Haruma
Giáo viên
8 tháng 7 2017
a)
Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)
\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)
\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)
b)
\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)
\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)
AH
Akai Haruma
Giáo viên
8 tháng 7 2017
c)
Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).
Đặt \(x+1=t\)
\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
PT
1
4 tháng 4 2016
\(=\frac{1}{2}\int\limits^1_0\ln\left(1+x^2\right)d\left(1+x^2\right)=\frac{1}{2}\left[\left(1+x^2\right)\ln\left(1+x^2\right)\right]|^1_0-\int\limits^1_0d\left(1+x^2\right)\)
\(=\frac{1}{2}\left[2\ln2-\left(1+x^2\right)|^1_0\right]=\frac{\left(2\ln2-1\right)}{2}\)
TT
1
4 tháng 4 2016
Đặt \(t=x^2+5\rightarrow\begin{cases}dt=2xdx,x=0\rightarrow t=5,x=3\rightarrow t=14\\f\left(x\right)dx=x\ln\left(x^2+5\right)dx=\frac{1}{2}\ln tdt\end{cases}\)
Do đó : \(I=\frac{1}{2}\int\limits^{14}_5\ln tdt=\frac{1}{2}\left(t\ln t\right)|^{14}_5=\frac{14\ln14-5\ln5-11}{2}\)
AH
Akai Haruma
Giáo viên
20 tháng 3 2017
Câu a)
\(I=\int ^{1}_{0}\frac{x(e^x+1)+1}{e^x+1}dx=\int ^{1}_{0}xdx+\int ^{1}_{0}\frac{dx}{e^x+1}\)
\(=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2}{2}+\int ^{1}_{0}\frac{d(e^x)}{e^x(e^x+1)}=\frac{1}{2}+\left.\begin{matrix} 1\\ 0\end{matrix}\right|\ln\left | \frac{e^x}{e^x+1} \right |\)
\(\Leftrightarrow I=\frac{3}{2}+\ln 2-\ln (e+1)\)
Câu d)
\(I=\int ^{e}_{1}\ln(x+1)d(x)=\int ^{e}_{1}\ln (x+1)d(x+1)\)
Đặt \(\left\{\begin{matrix} u=\ln (x+1)\\ dv=d(x+1)\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{d(x+1)}{x+1}\\ v=x+1\end{matrix}\right.\)
\(\Rightarrow I=\left.\begin{matrix} e\\ 1\end{matrix}\right|(x+1)\ln (x+1)-\int ^{e}_{1}d(x+1)\)
\(=(e+1)\ln \left ( \frac{e+1}{e} \right )-2\ln \left (\frac{2}{e}\right )\)
AH
Akai Haruma
Giáo viên
20 tháng 3 2017
Câu b)
Đặt \(\tan \frac{x}{2}=t\). Ta có:
\(\left\{\begin{matrix} dt=d\left ( \tan \frac{x}{2} \right )=\frac{1}{2\cos ^2\frac{x}{2}}dx=\frac{t^2+1}{2}dx\rightarrow dx=\frac{2dt}{t^2+1}\\\ \cos x=\frac{1-t^2}{t^2+1}\end{matrix}\right.\)
\( I=\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{1}{1+\cos x}dx}_{A}+\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{d(\cos x)}{\cos x+1}}_{B}\)
Có \(B=\int ^{\frac{\pi}{2}}_{0}\frac{d(\cos x+1)}{\cos x+1}=\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\ln |\cos x+1|=-\ln 2\)
\(A=\int ^{1}_{0}\frac{2dt}{(t^2+1)\frac{2}{t^2+1}}=\int ^{1}_{0}dt=1\)
\(\Rightarrow I=A+B=1-\ln 2\)
5 tháng 4 2016
Ta có \(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\ln2.\ln\left(2\tan x\right)}{\sin2x.\ln\left(2\tan x\right)}dx=\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}+\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}\)
Tính \(\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}=\frac{\ln2}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{d\left[\ln\left(2\tan x\right)\right]}{\ln2\left(2\tan x\right)}=\frac{\ln2}{2}\left[\ln\left(\ln\left(2\tan x\right)\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{\ln2}{2}.\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)\)
Tính \(\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}=\frac{1}{2}\ln\left(\tan x\right)|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{2}\ln\sqrt{3}\)
Vậy \(I=\frac{\ln2}{2}\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)+\frac{1}{2}\ln\sqrt{3}\)
Đặt \(u=\ln\left(x^2-x\right)\rightarrow du=\frac{2x-1}{x^2-x}dx,dv=dx\rightarrow v=x\)
Do đó : \(I=x.\ln\left(x^2-x\right)|^3_2-\int\limits^3_2\frac{x\left(2x-1\right)}{x\left(x-1\right)}dx=3\ln6-2\ln2-\int\limits^3_2\frac{2x-2+1}{x-1}dx\)
\(=\ln54-2\int\limits^3_2dx\frac{d\left(x-1\right)}{x-1}=\ln54-2-\ln\left(x-1\right)|^3_2=3\ln3-2\)