TG
1
K
Khách
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TV
1
1 tháng 4 2016
Đặt \(u=x^2\rightarrow du=2xdx,dv=\cos xdx\rightarrow v=\sin x\)
Do đó :
\(I=x^2.\sin x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_02x.\sin xdx=\frac{\pi^2}{4}+\int\limits^{\frac{\pi}{2}}_0x.d\left(\cos x\right)=\frac{\pi^2}{4}+\left(x.\cos x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\cos x\right)\)
\(=\frac{\pi^2}{4}+\left(0-\sin|^{\frac{\pi}{2}}_0\right)=\frac{\pi^2-4}{4}\)
LN
1
G
4 tháng 4 2016
\(I=\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\cos^2xdx=\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\left(\frac{1+\cos2x}{2}\right)dx=\int\limits^{\frac{\pi}{2}}_0\left(x-\frac{1}{2}\right)dx+\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\cos2xdx\)
\(=\left(\frac{1}{2}x^2-\frac{1}{2}x\right)|^{\frac{\pi}{2}}_0+\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)d\left(\sin2x\right)=\frac{\pi^2}{8}-\frac{\pi}{4}+\frac{1}{2}\left[\left(2x-1\right)\sin2x|^{\frac{\pi}{2}}_0-\int\limits^{^{\frac{\pi}{2}}_0}_0\sin2x.2dx\right]\)
\(=\frac{\pi^2}{8}-\frac{\pi}{4}+\left(0+\cos2x|^{\frac{\pi}{2}}_0\right)=\frac{\pi^2}{8}-\frac{\pi}{4}-1\)
NT
1
1 tháng 4 2016
\(\int\limits^{\frac{\pi}{4}}_0\frac{x}{\cos^2}dx=\int\limits^{\frac{\pi}{4}}_0x.d\left(\tan x\right)=x.\tan|^{\frac{\pi}{4}}_0-\int\limits^{\frac{\pi}{4}}_0\tan xdx=\frac{\pi}{4}+\ln\left(\cos x\right)|^{\frac{\pi}{4}}=\frac{\pi}{4}-\frac{1}{2}\ln2\)
LV
1
30 tháng 3 2016
\(I=-\frac{1}{2}\int_0^{\frac{\pi}{4}}\left(x^2-4x+3\right)d\cos2x\)
\(=-\frac{1}{2}\left[\left(x^2-4x+3\right)\cos2x\right]_0^{\frac{\pi}{4}}-\int^{^{\frac{\pi}{4}}}_0\cos2xd\left(x^2-4x+\right)\)
\(=\frac{3}{2}+\int^{^{\frac{\pi}{4}}}_0\left(x-2\right)\cos2xd=\frac{3}{2}+\frac{1}{2}\int^{^{\frac{\pi}{4}}}_0\left(x-2\right)\sin2x\)
\(=\frac{3}{2}+\frac{1}{2}\left[\left(x-2\right)\sin2x_0^{\frac{\pi}{4}}-\int^4_0\sin2dx\left(x-2\right)\right]\)
\(=\frac{3}{2}+\frac{1}{2}\left[\frac{\pi}{4}-2+\frac{1}{2}\cos2x_0^{\frac{\pi}{4}}\right]\)
\(=\frac{3}{2}+\frac{1}{2}\left[\frac{\pi}{4}-2-\frac{1}{2}\right]=\frac{\pi}{8}+\frac{1}{4}\)
PD
1
7 tháng 4 2016
\(I=\int\limits^{\pi}_0\left(x^2-x\sin x\right)dx=\frac{x^3}{3}|^{\pi}_0-\int^{\pi}_0x\sin xdx=\frac{\pi^3}{3}-\int\limits^{\pi}_0x\sin xdx\)
Tính \(I_1=\int\limits^{\pi}_0x\sin xdx\)
Đặt \(\begin{cases}u=x\\dv=\sin xdx\end{cases}\)\(\Rightarrow\begin{cases}du=dx\\v=-\cos x\end{cases}\)
\(\Rightarrow I_1=-x\cos x|^{\pi}_0+\int\limits^{\pi}_0\cos xdx=\pi+\sin x|^{\pi}_0=\pi\)
\(\Rightarrow I=\frac{\pi^3}{3}-\pi\)
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 \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
MT
2
1 tháng 4 2016
Đặt \(u=x^2e^x\Rightarrow du=\left(2x.e^x\right)dx=xe^x\left(2+x\right);dv=\frac{dx}{\left(x+2\right)^2}\Rightarrow v=-\frac{1}{x+2}\)
Vậy \(I=\int\limits^2_0\frac{x^2e^x}{\left(x+2\right)^2}=-\frac{x^2e^x}{x+2}|^2_0+\int\limits^2_0xe^xdx=-e^2+\left(xe^x-e\right)|^2=1_0\)
1 tháng 4 2016
Mình có cách khác, đổi biến số trước, sau lấy tích phân từng phần cũng ra
Đặt \(t=x+2\Rightarrow\begin{cases}dt=dx,x=0\Rightarrow t=2,x=2\rightarrow t=4\\f\left(x\right)dx=\frac{\left(t-2\right)^2e^{t-2}}{t}.dt=\left(t+\frac{2}{t}-4\right)e^{t-2}dt\end{cases}\)
Suy ra : \(I=\int\limits^4_2te^{t-2}dt+\int\limits^4_2\frac{e^{t-2}}{t}dt-4\int\limits^4_2e^{t-2}dt=J+K+4L\left(1\right)\)
Tính các tích phân J, K, L ta cũng ra được kết quả giống bạn Dương
\(\int\limits^{\frac{\pi}{2}}_0x.\sin^2xdx=\int\limits^{\frac{\pi}{2}}_0x\left(\frac{1-\cos2x}{2}\right)dx=\frac{1}{2}\left[\int\limits^{\frac{\pi}{2}}_0xdx-\int\limits^{\frac{\pi}{2}}_0x.\cos3xdx\right]\)
\(=\frac{1}{2}\left(\frac{1}{2}x^2|^{\frac{\pi}{2}}_0-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0x.d\left(\sin2x\right)\right)\)
\(=\frac{1}{2}\left[\frac{\pi^2}{8}-\frac{1}{2}\left(x.\sin2x\right)|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\sin2xdx\right]\)
\(=\frac{1}{2}\left[\frac{\pi^2}{8}-\frac{1}{2}\left(0+\frac{1}{2}\cos2x|^{\frac{\pi}{2}}_0\right)\right]=\frac{\pi^2+8}{16}\)