Question Number 84862 by Jakir Sarif Mondal last updated on 25/Mar/20
![∫_( 0) ^(100) [tan^(−1) x]dx =? −Jakir Sarif Mondal.](https://www.tinkutara.com/question/Q84862.png)
$$\underset{\:\mathrm{0}} {\overset{\mathrm{100}} {\int}}\:\left[\mathrm{tan}^{−\mathrm{1}} {x}\right]{dx}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\boldsymbol{{Jakir}}\:\boldsymbol{{Sarif}}\:\:\boldsymbol{{Mondal}}. \\ $$$$\:\:\:\:\:\:\: \\ $$
Commented by mr W last updated on 17/Mar/20
![for x≥0 the range of tan^(−1) (x) is [0,(π/2)). for 0≤x<tan 1: 0≤tan^(−1) x<1 ⇒[tan^(−1) x]=0 for tan 1≤x<+∞: 1≤tan^(−1) x<(π/2)<2 ⇒[tan^(−1) x]=1 ∫_0 ^(100) [tan^(−1) x]dx =∫_0 ^(tan 1) [tan^(−1) x]dx+∫_(tan 1) ^(100) [tan^(−1) x]dx =∫_0 ^(tan 1) 0 dx+∫_(tan 1) ^(100) 1 dx =0+(100−tan 1) =100−tan 1](https://www.tinkutara.com/question/Q84885.png)
$${for}\:{x}\geqslant\mathrm{0}\:{the}\:{range}\:{of}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:{is}\:\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right). \\ $$$${for}\:\mathrm{0}\leqslant{x}<\mathrm{tan}\:\mathrm{1}:\:\mathrm{0}\leqslant\mathrm{tan}^{−\mathrm{1}} {x}<\mathrm{1}\:\Rightarrow\left[\mathrm{tan}^{−\mathrm{1}} {x}\right]=\mathrm{0} \\ $$$${for}\:\mathrm{tan}\:\mathrm{1}\leqslant{x}<+\infty:\:\mathrm{1}\leqslant\mathrm{tan}^{−\mathrm{1}} {x}<\frac{\pi}{\mathrm{2}}<\mathrm{2}\:\Rightarrow\left[\mathrm{tan}^{−\mathrm{1}} {x}\right]=\mathrm{1} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{100}} \left[\mathrm{tan}^{−\mathrm{1}} {x}\right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{tan}\:\mathrm{1}} \left[\mathrm{tan}^{−\mathrm{1}} {x}\right]{dx}+\int_{\mathrm{tan}\:\mathrm{1}} ^{\mathrm{100}} \left[\mathrm{tan}^{−\mathrm{1}} {x}\right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{tan}\:\mathrm{1}} \mathrm{0}\:{dx}+\int_{\mathrm{tan}\:\mathrm{1}} ^{\mathrm{100}} \mathrm{1}\:{dx} \\ $$$$=\mathrm{0}+\left(\mathrm{100}−\mathrm{tan}\:\mathrm{1}\right) \\ $$$$=\mathrm{100}−\mathrm{tan}\:\mathrm{1} \\ $$
Answered by Rio Michael last updated on 16/Mar/20
![let u = tan^(−1) x and (dv/dx) = 1 (du/dx) = (1/(1 + x^2 )) and v = x ⇒ ∫_0 ^(100) tan^(−1) xdx= [xtan^(−1) x]_0 ^(100) − ∫_0 ^(100) (x/(1 + x^2 )) dx = 100 tan^(−1) 100 − (1/2)ln(1 + x^2 )∣_0 ^(100) = 100tan^(−1) 100 −(1/2)ln(1 + 100^2 )](https://www.tinkutara.com/question/Q84863.png)
$$\:\mathrm{let}\:{u}\:=\:\mathrm{tan}^{−\mathrm{1}} {x}\:\:\mathrm{and}\:\frac{{dv}}{{dx}}\:=\:\mathrm{1} \\ $$$$\:\:\:\frac{{du}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:\mathrm{and}\:{v}\:=\:{x} \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{100}} {\int}}\mathrm{tan}^{−\mathrm{1}} {xdx}=\:\left[{x}\mathrm{tan}^{−\mathrm{1}} {x}\right]_{\mathrm{0}} ^{\mathrm{100}} −\:\underset{\mathrm{0}} {\overset{\mathrm{100}} {\int}}\frac{{x}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{100}\:\mathrm{tan}^{−\mathrm{1}} \mathrm{100}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\underset{\mathrm{0}} {\overset{\mathrm{100}} {\mid}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{100tan}^{−\mathrm{1}} \mathrm{100}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}\:+\:\mathrm{100}^{\mathrm{2}} \right) \\ $$
Answered by TANMAY PANACEA last updated on 17/Mar/20
![tan^(−1) x=1 →x=tan1 ∫_0 ^(tan1) [tan^(−1) x]dx+∫_(tan1) ^(100) [tan^(−1) x]dx =∫_0 ^(tan1) 0×dx+1×∣x∣_(tan1) ^(100) =(100−tan1)](https://www.tinkutara.com/question/Q84877.png)
$${tan}^{−\mathrm{1}} {x}=\mathrm{1}\:\rightarrow{x}={tan}\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{{tan}\mathrm{1}} \left[{tan}^{−\mathrm{1}} {x}\right]{dx}+\int_{{tan}\mathrm{1}} ^{\mathrm{100}} \left[{tan}^{−\mathrm{1}} {x}\right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{{tan}\mathrm{1}} \mathrm{0}×{dx}+\mathrm{1}×\mid{x}\mid_{{tan}\mathrm{1}} ^{\mathrm{100}} \\ $$$$=\left(\mathrm{100}−{tan}\mathrm{1}\right) \\ $$