Integral Calculus

Vishal U. Karve's Avatar
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20 Feb 2012 15:34:56 IST
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Integration Challenge-4!!
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Integral Calculus

Integration Challenge-4!!

              

Integrate:

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jagdish singh's Avatar

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20 Feb 2012 17:39:15 IST
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 [img]http://latex.codecogs.com/gif.latex?\hspace{-16}\mathbf{\int\frac{x^2}{(x\cos%20x-\sin%20x).(x\sin%20x+\cos%20x)}dx}$\\\\\\%20Now%20We%20Can%20Write%20$\mathbf{x^2=x^2(\sin^2%20x+\cos^2%20x)}$\\\\\\%20So%20$\mathbf{x^2=(x\cos%20x-\sin%20x).\frac{d}{dx}(x\sin%20x+\cos%20x)}$\\\\\\%20$\mathbf{-(x\sin%20x+cos%20x).\frac{d}{dx}(x\cos%20x-\sin%20x)}$\\\\\\%20So%20$\mathbf{x^2=(x\cos%20x-\sin%20x).(x\cos%20x)-(x\cos%20x+\sin%20x)(-x\sin%20x)}$\\\\\\%20So%20Original%20Integral%20Integral%20Convert%20into%20\\\\\\%20$\mathbf{=\int\frac{(x\cos%20x-\sin%20x).(x\cos%20x)-(x\cos%20x+\sin%20x)(-x\sin%20x)}{(x\cos%20x-\sin%20x).(x\sin%20x+\cos%20x)}}$\\\\\\%20$\mathbf{=\ln%20\left|x\sin%20x+\cos%20x%20\right%20|-\ln\left|x\cos%20x-\sin%20x\right|+C}$\\\\\\%20So%20$\mathbf{\int\frac{x^2}{(x\cos%20x-\sin%20x).(x\sin%20x+\cos%20x)}dx=\ln%20\left|\frac{x\sin%20x+\cos%20x%20}{x\cos%20x-\sin%20x}\right|+C}$[img]

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20 Feb 2012 17:39:58 IST
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