Tính tích phân :
\(I=\int^{\frac{\pi}{4}}_0\left(x^2-4x+3\right)\sin2xdx\)
Tính tích phân :
\(I=\int^{\frac{\pi}{4}}_0\left(x^2-4x+3\right)\sin2xdx\)
\(I=-\frac{1}{2}\int_0^{\frac{\pi}{4}}\left(x^2-4x+3\right)d\cos2x\)
\(=-\frac{1}{2}\left[\left(x^2-4x+3\right)\cos2x\right]_0^{\frac{\pi}{4}}-\int^{^{\frac{\pi}{4}}}_0\cos2xd\left(x^2-4x+\right)\)
\(=\frac{3}{2}+\int^{^{\frac{\pi}{4}}}_0\left(x-2\right)\cos2xd=\frac{3}{2}+\frac{1}{2}\int^{^{\frac{\pi}{4}}}_0\left(x-2\right)\sin2x\)
\(=\frac{3}{2}+\frac{1}{2}\left[\left(x-2\right)\sin2x_0^{\frac{\pi}{4}}-\int^4_0\sin2dx\left(x-2\right)\right]\)
\(=\frac{3}{2}+\frac{1}{2}\left[\frac{\pi}{4}-2+\frac{1}{2}\cos2x_0^{\frac{\pi}{4}}\right]\)
\(=\frac{3}{2}+\frac{1}{2}\left[\frac{\pi}{4}-2-\frac{1}{2}\right]=\frac{\pi}{8}+\frac{1}{4}\)
Tính tích phân :
\(\int^1_0\left(\frac{x^2-4x+3}{e^{2x}}\right)dx\)
Đặt \(u=\left(x^3-2x^x+3x+1\right)\Rightarrow du=\left(3x^2-4x+3\right)dx;dv=\frac{dx}{e^{2x}}\Rightarrow v=-\frac{2}{e^{2x}}\)
Ta được : \(-\frac{2}{e^{2x}}\left(x^3-2x^2+3x+1\right)|^1_0+2\int\limits^1_0\left(\frac{3x^2-4x+3}{e^{2x}}\right)dx=2-\frac{6}{e^2}+2J\)
Tương tự ta tính J
Đăth \(u_1=\left(3x^2-4x+3\right)\Rightarrow du_1=\left(6x-4\right)dx;dv_1=\frac{dx}{e^{2x}}\Rightarrow v_1=-\frac{2}{e^{2x}}\left(1\right)\)
Do đó :
\(J=-\frac{2}{e^{2x}}\left(3x^2-4x+3\right)|^1_0+2\int\limits^1_0\frac{6x-4}{e^{2x}}dx=6-\frac{4}{e^2}+2K\left(2\right)\)
Ta tính K :
\(K=\int\limits^1_0\frac{6x-4}{e^{2x}}dx\)
Đặt \(u_2=6x-4\Rightarrow du_2=6dx;dv_2=\frac{dx}{e^{2x}}\Rightarrow v_2=-\frac{2}{e^{2x}}\)
Do đó : \(K=-\frac{2}{e^{2x}}\left(x-4\right)|^1_0+2\int\limits^1_0\frac{6dx}{e^{2x}}=\frac{6}{e^x}-8-6\frac{1}{e^{2x}}|^1_0\left(\frac{1}{e^2}-1\right)=-2\left(3\right)\)
Thay (3) vào (2)
\(J=6-\frac{4}{e^2}+2\left(-2\right)=2-\frac{4}{e^2}\)
Lại thay vào (1) ta có :
\(I=2-\frac{6}{e^2}+2\left(2-\frac{4}{e^2}\right)=6-\frac{14}{e^2}\)
Tính tích phân :
\(\int^1_0x^3e^{x^2}dx\)
\(\int\limits^1_0x^3e^{x^2}dx=\int\limits^1_0x^3e^{x^2}.xdx\)
Đặt \(t=x^2\Rightarrow\begin{cases}dt=2xdx;x=0\rightarrow t=0,x=1\rightarrow t=1\\f\left(x\right)dx=te^tdt\end{cases}\)
Do đó : \(I=\int\limits^1_0te^1dt=\frac{1}{2}\int\limits^1_0t.d\left(e^t\right)=\frac{1}{2}\left(t.e^t-e^t\right)|^1_0=\frac{1}{2}\)
Tính tích phân :
\(\int^2_0\frac{x^2e^x}{\left(x+2\right)^2}dx\)
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 \(I=\int\limits^2_0\frac{x^2e^x}{\left(x+2\right)^2}=-\frac{x^2e^x}{x+2}|^2_0+\int\limits^2_0xe^xdx=-e^2+\left(xe^x-e\right)|^2=1_0\)
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
Tính tích phân :
\(\int^{\frac{\pi}{2}}_0x\sin^2xdx\)
\(\int\limits^{\frac{\pi}{2}}_0x.\sin^2xdx=\int\limits^{\frac{\pi}{2}}_0x\left(\frac{1-\cos2x}{2}\right)dx=\frac{1}{2}\left[\int\limits^{\frac{\pi}{2}}_0xdx-\int\limits^{\frac{\pi}{2}}_0x.\cos3xdx\right]\)
\(=\frac{1}{2}\left(\frac{1}{2}x^2|^{\frac{\pi}{2}}_0-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0x.d\left(\sin2x\right)\right)\)
\(=\frac{1}{2}\left[\frac{\pi^2}{8}-\frac{1}{2}\left(x.\sin2x\right)|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\sin2xdx\right]\)
\(=\frac{1}{2}\left[\frac{\pi^2}{8}-\frac{1}{2}\left(0+\frac{1}{2}\cos2x|^{\frac{\pi}{2}}_0\right)\right]=\frac{\pi^2+8}{16}\)
Tính tích phân :
\(\int^{\frac{\pi}{2}}_0x^2\cos xdx\)
Đặt \(u=x^2\rightarrow du=2xdx,dv=\cos xdx\rightarrow v=\sin x\)
Do đó :
\(I=x^2.\sin x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_02x.\sin xdx=\frac{\pi^2}{4}+\int\limits^{\frac{\pi}{2}}_0x.d\left(\cos x\right)=\frac{\pi^2}{4}+\left(x.\cos x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\cos x\right)\)
\(=\frac{\pi^2}{4}+\left(0-\sin|^{\frac{\pi}{2}}_0\right)=\frac{\pi^2-4}{4}\)
Tính tích phân :
\(\int^{\frac{\pi}{4}}_0\frac{x}{\cos^2}dx\)
\(\int\limits^{\frac{\pi}{4}}_0\frac{x}{\cos^2}dx=\int\limits^{\frac{\pi}{4}}_0x.d\left(\tan x\right)=x.\tan|^{\frac{\pi}{4}}_0-\int\limits^{\frac{\pi}{4}}_0\tan xdx=\frac{\pi}{4}+\ln\left(\cos x\right)|^{\frac{\pi}{4}}=\frac{\pi}{4}-\frac{1}{2}\ln2\)
Tính tích phân :
\(\int\limits^e_1x^3\ln^2xdx\)
Đặt \(u=\ln^2x\rightarrow du=2\ln x\frac{dx}{x},dv=\int\limits x^3dx\rightarrow v=\frac{1}{4}x^4\)
Do đó : \(I=\frac{1}{4}x^4.\ln^2x|^e_1-\frac{1}{4}\int\limits^e_12\ln x.\frac{x^4}{x}dx=\frac{e^4}{4}-\frac{1}{2}\int\limits^e_1x^3\ln sdx=\frac{e^4}{4}-\frac{1}{2}J\left(1\right)\)
Tính \(J=\int\limits^e_1x^3\ln xdx\)
Đặt \(u_1=\ln x\rightarrow du_1=\frac{dx}{x},dv_1=\int x^3dx\rightarrow v_1=\frac{1}{4}x^4\)
Do đó :
\(J=\frac{1}{4}x^4\ln x|^e_1-\frac{1}{4}\int\limits^e_1x^3dx=\frac{e^4}{4}-\frac{1}{16}x^2|^e_1=\frac{3e^4+1}{16}\)
Thay vào (1) ta có :
\(I=\frac{e^4}{4}-\frac{1}{2}\left(\frac{3e^4+1}{16}\right)=\frac{5e^4-1}{32}\)
Tính tích phân :
\(\int\limits^2_3\ln\left(x^2-x\right)dx\)
Đặt \(u=\ln\left(x^2-x\right)\rightarrow du=\frac{2x-1}{x^2-x}dx,dv=dx\rightarrow v=x\)
Do đó : \(I=x.\ln\left(x^2-x\right)|^3_2-\int\limits^3_2\frac{x\left(2x-1\right)}{x\left(x-1\right)}dx=3\ln6-2\ln2-\int\limits^3_2\frac{2x-2+1}{x-1}dx\)
\(=\ln54-2\int\limits^3_2dx\frac{d\left(x-1\right)}{x-1}=\ln54-2-\ln\left(x-1\right)|^3_2=3\ln3-2\)
Tính tích phân :
\(\int\limits^e_1\ln^3xdx\)
Đặt \(u=\ln^3x\rightarrow du=3\ln^2x\frac{dx}{x},dv=dx\rightarrow v=x\)
Do đó : \(I=x\ln^3x|^e_1-3\int\limits^3_1\ln^2xdx=e-3J\left(1\right)\)
Tính \(J=\int\limits^e_1\ln^2xdx\)
Đặt \(u_1=\ln^2x\rightarrow du_1=\frac{2\ln x}{x}dx,dv_1=dx\rightarrow v_1=x\)
Do vậy, \(J=x\ln^2x|^e_1-2\int\limits^e_1\ln xdx=e-2\left(x\ln x|^e_1-\int\limits^e_1dx\right)=e-2\left(x\ln x-x\right)|^e_1=e-2\)
Thay vào (1) ta có : \(I=e-3\left(e-2\right)=6-2e\)