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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 :
\(I=x^2.\sin x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_02x.\sin xdx=\frac{\pi^2}{4}+\int\limits^{\frac{\pi}{2}}_0x.d\left(\cos x\right)=\frac{\pi^2}{4}+\left(x.\cos x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\cos x\right)\)
\(=\frac{\pi^2}{4}+\left(0-\sin|^{\frac{\pi}{2}}_0\right)=\frac{\pi^2-4}{4}\)
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 \(\begin{cases}u=x\\dv=\sin xdx\end{cases}\)\(\Rightarrow\begin{cases}du=dx\\v=-\cos x\end{cases}\)
\(\Rightarrow I_1=-x\cos x|^{\pi}_0+\int\limits^{\pi}_0\cos xdx=\pi+\sin x|^{\pi}_0=\pi\)
\(\Rightarrow I=\frac{\pi^3}{3}-\pi\)
LN
1
G
4 tháng 4 2016
\(I=\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\cos^2xdx=\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\left(\frac{1+\cos2x}{2}\right)dx=\int\limits^{\frac{\pi}{2}}_0\left(x-\frac{1}{2}\right)dx+\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\cos2xdx\)
\(=\left(\frac{1}{2}x^2-\frac{1}{2}x\right)|^{\frac{\pi}{2}}_0+\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)d\left(\sin2x\right)=\frac{\pi^2}{8}-\frac{\pi}{4}+\frac{1}{2}\left[\left(2x-1\right)\sin2x|^{\frac{\pi}{2}}_0-\int\limits^{^{\frac{\pi}{2}}_0}_0\sin2x.2dx\right]\)
\(=\frac{\pi^2}{8}-\frac{\pi}{4}+\left(0+\cos2x|^{\frac{\pi}{2}}_0\right)=\frac{\pi^2}{8}-\frac{\pi}{4}-1\)
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 \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
NH
1
11 tháng 4 2016
\(I=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\cos2x+3\cos x+2}dx=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{2\cos^2x+3\cos x+1}dx\)
Đặt \(\cos x=t\Rightarrow dt=-\sin dx\)
Với \(x=0\Rightarrow t=1\)
Với \(x=\frac{\pi}{2}\Rightarrow t=0\)
\(I=\int\limits^1_0\frac{dt}{2t^2+3t+1}=\int\limits^1_0\frac{dt}{\left(2t+1\right)\left(t+1\right)}=2\int\limits^1_0\left(\frac{1}{2t+1}+\frac{1}{2t+1}\right)dt\)
\(=\left(\ln\frac{2t+1}{2t+1}\right)|^1_0=\ln\frac{3}{2}\)
HT
20 tháng 3 2021
Cách này hơi dài chút, nhưng nếu nghĩ ra cách hay hơn mình sẽ đề xuất nhe!
\(=\int\sin^5x.\left(2\sin x\cos x\right)^3.2xdx=16\int x.\sin^8x\cos^3xdx\)
\(\left\{{}\begin{matrix}u=x\\dv=\sin^8x.\cos^3xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\int\sin^8x.\cos^3xdx\end{matrix}\right.\)
\(I_1=\int\sin^8x\cos^3xdx=\int\sin^8x.\cos^2x.\cos xdx=\int\sin^8x.\left(1-\sin^2x\right)\cos xdx\)
\(t=\sin x\Rightarrow dt=\cos xdx\Rightarrow\int\sin^8x\left(1-\sin^2x\right)\cos xdx=\int(t^8-t^{10})dt=\dfrac{1}{9}t^9-\dfrac{1}{11}t^{11}=\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\)
\(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\end{matrix}\right.\)
\(\Rightarrow\dfrac{I}{16}=x.\left(\dfrac{1}{9}\sin^9x-11\sin^{11}x\right)-\int\left(\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\right)dx\)
\(I_2=\int\left(\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\right)dx=\dfrac{1}{9}\int\sin^9xdx-\dfrac{1}{11}\int\sin^{11}xdx\)
À thế này là xong rồi còn gì :) Bạn tự làm nốt nhé
\(\int\limits^{\frac{\pi}{2}}_0x.\sin^2xdx=\int\limits^{\frac{\pi}{2}}_0x\left(\frac{1-\cos2x}{2}\right)dx=\frac{1}{2}\left[\int\limits^{\frac{\pi}{2}}_0xdx-\int\limits^{\frac{\pi}{2}}_0x.\cos3xdx\right]\)
\(=\frac{1}{2}\left(\frac{1}{2}x^2|^{\frac{\pi}{2}}_0-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0x.d\left(\sin2x\right)\right)\)
\(=\frac{1}{2}\left[\frac{\pi^2}{8}-\frac{1}{2}\left(x.\sin2x\right)|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\sin2xdx\right]\)
\(=\frac{1}{2}\left[\frac{\pi^2}{8}-\frac{1}{2}\left(0+\frac{1}{2}\cos2x|^{\frac{\pi}{2}}_0\right)\right]=\frac{\pi^2+8}{16}\)