defi intregation, the value of the constant a>0 such that integation from 0 to a[tan<sup>-1</sup>x^1/2]dx=intega

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Title: defi intregation

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Posted On: Feb 06, 2012 10:18 PM

the value of the constant a>0 such that integation from 0 to a[tan^-1x^1/2]dx=integation from 0 to a [cot^-1x^1/2]dx,where [.] denotes the greatest integer function.

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debadutta mishra
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jagdish singh

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Feb 07, 2012 10:20 PM

$\hspace{-16}\mathbf{\int_{0}^{a}\left[\tan^{-1}\sqrt{x}\right]dx=\int_{0}^{a}\left[\cot^{-1}\sqrt{x}\right]dx}$\\\\\\ Now Here we have to find the value of $\mathbf{x}$ for which\\\\\\ $\mathbf{\left[\tan^{-1}\sqrt{x}\right]}$ and $\mathbf{\left[\cot^{-1}\sqrt{x}\right]}$ is an Integer.So\\\\\\ $\mathbf{\int_{0}^{\tan^2 (1)}\left[\tan^{-1}\sqrt{x}\right]dx+\int_{\tan^2 (1)}^{a}\left[\tan^{-1}\sqrt{x}\right]dx}$\\\\\\ $\mathbf{=\int_{0}^{\cot^2(1)}\left[\cot^{-1}\sqrt{x}\right]dx+\int_{\cot^2(1)}^{a}\left[\cot^{-1}\sqrt{x}\right]dx}$\\\\\\ So $\mathbf{\int_{0}^{\tan^2(1)}0.dx+\int_{\tan^2(1)}^{a}1.dx=\int_{0}^{\cot^2(1)}1.dx+\int_{\cot^2(1)}^{a}0.dx}$\\\\\\ So $\mathbf{a-\tan^2(1)=\cot^2(1)}$\\\\\\ So $\mathbf{a=\tan^2(1)+\cot^2(1)}$$

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defi intregationin	integration
indefinite integrals	definite integral
definite integral	question-differential eqns
limits	limits
integration	Indefinite integration