\(\sin2x\)\(-\)(\(\sin x-\cos x-1\))(2\(\times\sin x-\cos x-3\))\(=0\)
giai giup minh voi
\(\sin2x\)\(-\)(\(\sin x-\cos x-1\))(2\(\times\sin x-\cos x-3\))\(=0\)
giai giup minh voi
\(sin 2x-(2sin^2 x-sin2x-2sinx-1/2.\sin 2x+\cos^2x+\cos x-3\sin x-3\cos x+3)=0\)
\(5\sin x.\cos x+5\sin x+2\cos x-\sin^2x-4=0\)
\(\cos x(5\sin x+2)=\sin^2x-5\sin x+4=(\sin x-1)(\sin x -4)\)
Bình phương 2 vế suy ra
\((1-\sin^2 x)(5\sin x+2)^2=(1-\sin x)^2(\sin x-4)^2\)
TH1: \(\sin x=1\)
TH 2: \((1+\sin x)(5\sin x+2)^2=(1-\sin x)(\sin x-4)^2\)
Tính cá tích phân sau:
I = \(\int\limits_0^1 {x^2\over \sqrt{3+2x-x^2}}dx\)
I = \(\int\limits_1^\sqrt2 {\sqrt{x^2-1}\over x}dx\)
I = \(\int\limits_1^2 {x+1\over \sqrt{x(2-x)}}dx\)
I = \(\int\limits_0^1 {dx\over x^2+x+1}\)
Cậu sống ở đâu hở ? Lấy đâu ra toán khó thế ?
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 : \(I=e-3\left(e-2\right)=6-2e\)
Tính tích phân :
\(\int\limits^e_1x^2\ln xdx\)
Đặt \(u=\ln x\rightarrow du=\frac{dx}{x};dv=\int x^2dx\rightarrow v=\frac{1}{3}x^3\)
Do đó : \(I=\frac{1}{3}x^3\ln x|^e_1-\frac{1}{3}\int\limits^e_1x^2dx=\frac{e^3}{3}-\frac{1}{3}x^3|^e_1=\frac{2e^3+1}{9}\)
Tính tích phân :
\(I=\int\limits^3_1\frac{2}{2x^2+3x-2}dx\)
Ta có : \(I=\int\limits^3_1\frac{2}{\left(2x-1\right)\left(x+2\right)}dx=\frac{2}{5}\left(\int\limits^3_1\frac{2}{2x-1}-\int\limits^3_1\frac{1}{x+2}dx\right)\)
\(=\frac{2}{5}\left(\int\limits^3_1\frac{d\left(2x-1\right)}{2x-1}-\int\limits^3_1\frac{d\left(x+2\right)}{x+2}\right)\)
\(=\frac{2}{5}\left(\ln\left|2x-1\right||^3_1-\ln\left|x+2\right||^3_1\right)=\frac{2}{5}\ln3\)
Tính tích phân :
\(I=\int\limits_1^2\left(2x^2+\ln x\right)dx\)
Ta có : \(I=\int\limits^2_12x^3dx+\int\limits^2_1\ln xdx\)
Đặt \(I_1=\int\limits^2_12x^3dx\) và \(I_2=\int\limits^2_1\ln xdx\)
Ta có :
\(I_1=\frac{1}{2}x^4|^2_1=\frac{15}{2}\)
\(I_2=x.\ln x|^2_1-\int_1xd^2\left(\ln x\right)=2\ln2-x|^2_1=2\ln2-1\)
Vậy \(I=I_1+I_2=\frac{13}{2}+2\ln2\)