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Question Number 84862 by Jakir Sarif Mondal last updated on 25/Mar/20

$$\underset{0}{\stackrel{100}{\int }}\left[{\tan }^{-1}x\right]dx=?$$$$-\mathit{J}\mathit{a}\mathit{k}\mathit{i}\mathit{r}\mathit{S}\mathit{a}\mathit{r}\mathit{i}\mathit{f}\mathit{M}\mathit{o}\mathit{n}\mathit{d}\mathit{a}\mathit{l}.$$$$$$

Commented by mr W last updated on 17/Mar/20

$$forx⩾0therangeof{\tan }^{-1}\left(x\right)is[0,\frac{\pi }{2}).$$$$for0⩽x<\tan 1:0⩽{\tan }^{-1}x<1\Rightarrow \left[{\tan }^{-1}x\right]=0$$$$for\tan 1⩽x<+\mathrm{\infty }:1⩽{\tan }^{-1}x<\frac{\pi }{2}<2\Rightarrow \left[{\tan }^{-1}x\right]=1$$$$$$$${\int }_0^{100}\left[{\tan }^{-1}x\right]dx$$$$={\int }_0^{\tan 1}\left[{\tan }^{-1}x\right]dx+{\int }_{\tan 1}^{100}\left[{\tan }^{-1}x\right]dx$$$$={\int }_0^{\tan 1}0dx+{\int }_{\tan 1}^{100}1dx$$$$=0+(100-\tan 1)$$$$=100-\tan 1$$

Answered by Rio Michael last updated on 16/Mar/20

$$\mathrm{let}u={\tan }^{-1}x\mathrm{and}\frac{dv}{dx}=1$$$$\frac{du}{dx}=\frac{1}{1+x^2}\mathrm{and}v=x$$$$\Rightarrow \underset{0}{\stackrel{100}{\int }}{\tan }^{-1}xdx={\left[x{\tan }^{-1}x\right]}_0^{100}-\underset{0}{\stackrel{100}{\int }}\frac{x}{1+x^2}dx$$$$=100{\tan }^{-1}100-\frac{1}{2}\ln (1+x^2)\underset{0}{\stackrel{100}{\mid }}$$$$={100\tan }^{-1}100-\frac{1}{2}\ln (1+{100}^2)$$

Answered by TANMAY PANACEA last updated on 17/Mar/20

$${tan}^{-1}x=1\to x=tan1$$$${\int }_0^{tan1}\left[{tan}^{-1}x\right]dx+{\int }_{tan1}^{100}\left[{tan}^{-1}x\right]dx$$$$={\int }_0^{tan1}0\times dx+1\times \mid x{\mid }_{tan1}^{100}$$$$=(100-tan1)$$

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