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TG
1
1 tháng 4 2016
\(\int\limits^{\frac{\pi}{2}}_0x.\sin^2xdx=\int\limits^{\frac{\pi}{2}}_0x\left(\frac{1-\cos2x}{2}\right)dx=\frac{1}{2}\left[\int\limits^{\frac{\pi}{2}}_0xdx-\int\limits^{\frac{\pi}{2}}_0x.\cos3xdx\right]\)
\(=\frac{1}{2}\left(\frac{1}{2}x^2|^{\frac{\pi}{2}}_0-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0x.d\left(\sin2x\right)\right)\)
\(=\frac{1}{2}\left[\frac{\pi^2}{8}-\frac{1}{2}\left(x.\sin2x\right)|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\sin2xdx\right]\)
\(=\frac{1}{2}\left[\frac{\pi^2}{8}-\frac{1}{2}\left(0+\frac{1}{2}\cos2x|^{\frac{\pi}{2}}_0\right)\right]=\frac{\pi^2+8}{16}\)
NT
1
1 tháng 4 2016
\(\int\limits^{\frac{\pi}{4}}_0\frac{x}{\cos^2}dx=\int\limits^{\frac{\pi}{4}}_0x.d\left(\tan x\right)=x.\tan|^{\frac{\pi}{4}}_0-\int\limits^{\frac{\pi}{4}}_0\tan xdx=\frac{\pi}{4}+\ln\left(\cos x\right)|^{\frac{\pi}{4}}=\frac{\pi}{4}-\frac{1}{2}\ln2\)
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
4 tháng 4 2016
\(I=\int\limits^{\ln3}_1\left(x^2-2x\right)de^x=\left(x^2-2x\right)e^x|^{\ln3}_1-\int\limits_1^{\ln3}e^xd\left(x^2-2x\right)=3\left(\ln^23-2\ln3\right)+e-2\int\limits^{\ln3}_1\left(x-1\right)e^xdx\)
\(\int\limits^{\ln3}_1\left(x-1\right)e^xdx=k\)
Lại có :
\(k=\int\limits^{\ln3}_1\left(x-1\right)de^x=\left(x-1\right)e^x|^{\ln3}_0-\int\limits^{\ln3}_0e^xd\left(x-1\right)=3\left(\ln3-1\right)-e^x|^{\ln3}_0=3\ln3-6+e\)
Do đó :
\(I=3\left(\ln^23-2\ln3\right)+e-2\left(3\ln3-6+e\right)=3\ln^23-12\ln3+12-e\)
LV
1
30 tháng 3 2016
\(I=-\frac{1}{2}\int_0^{\frac{\pi}{4}}\left(x^2-4x+3\right)d\cos2x\)
\(=-\frac{1}{2}\left[\left(x^2-4x+3\right)\cos2x\right]_0^{\frac{\pi}{4}}-\int^{^{\frac{\pi}{4}}}_0\cos2xd\left(x^2-4x+\right)\)
\(=\frac{3}{2}+\int^{^{\frac{\pi}{4}}}_0\left(x-2\right)\cos2xd=\frac{3}{2}+\frac{1}{2}\int^{^{\frac{\pi}{4}}}_0\left(x-2\right)\sin2x\)
\(=\frac{3}{2}+\frac{1}{2}\left[\left(x-2\right)\sin2x_0^{\frac{\pi}{4}}-\int^4_0\sin2dx\left(x-2\right)\right]\)
\(=\frac{3}{2}+\frac{1}{2}\left[\frac{\pi}{4}-2+\frac{1}{2}\cos2x_0^{\frac{\pi}{4}}\right]\)
\(=\frac{3}{2}+\frac{1}{2}\left[\frac{\pi}{4}-2-\frac{1}{2}\right]=\frac{\pi}{8}+\frac{1}{4}\)
Đặt \(u=x^2\rightarrow du=2xdx,dv=\cos xdx\rightarrow v=\sin x\)
Do đó :
\(I=x^2.\sin x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_02x.\sin xdx=\frac{\pi^2}{4}+\int\limits^{\frac{\pi}{2}}_0x.d\left(\cos x\right)=\frac{\pi^2}{4}+\left(x.\cos x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\cos x\right)\)
\(=\frac{\pi^2}{4}+\left(0-\sin|^{\frac{\pi}{2}}_0\right)=\frac{\pi^2-4}{4}\)