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5 Feb 2012 14:40:47 IST
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integration of 1/(sinx+secx)
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5 Feb 2012 16:05:37 IST
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Jagdish Singh.(Pantnagar) |
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6 Feb 2012 15:39:19 IST
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indefinite integration
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6 Feb 2012 17:04:16 IST
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dx=dt$\\\\\\\\%20$\int%20\frac{1}{(\sqrt{3})^2-t^2}dt=\frac{1}{2\sqrt{3}}\ln\left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+C_{1}$\\\\\\\\%20Similarly%20$I_{2}=\frac{1}{2}.\int\frac{\cos%20x-\sin%20x}{\sin%20x.\cos%20x+1}dx$\\\\\\\\%20Let%20$\sin%20x+\cos%20x=u\Leftrightarrow%20(\cos%20x-\sin%20x)dx=du$\\\\\\\\%20So%20$I_{2}=\frac{1}{u^2+1}du=\tan^{-1}(u)+C_{2}$\\\\\\\\%20$\int\frac{1}{\sin%20x+\sec%20x}dx=\frac{1}{2\sqrt{3}}\ln\left|\frac{\sin%20x-\cos%20x-\sqrt{3}}{\sin%20x-\cos%20x+\sqrt{3}}\right|+\tan^{-1}\left(\sin%20x+\cos%20x\right)+C$)
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Jagdish Singh.(Pantnagar) |
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