Question Number 61614 by Tajaddin last updated on 05/Jun/19
$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\:\frac{{d}}{{dx}}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right){dx}\:= \\ $$
Commented by maxmathsup by imad last updated on 05/Jun/19
![we have (d/dx)(arctan((1/x))) =((−1)/(x^2 (1+(1/x^2 )))) =((−1)/(x^2 +1)) ⇒ ∫_(−1) ^1 (d/dx)(arctan((1/x)))dx =−∫_(−1) ^1 (dx/(1+x^2 )) =−2 ∫_0 ^1 (dx/(1+x^2 )) =−2[arctanx]_0 ^1 =−2(π/4) =−(π/2)](https://www.tinkutara.com/question/Q61628.png)
$${we}\:{have}\:\frac{{d}}{{dx}}\left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)\:=\frac{−\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:=\frac{−\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{{d}}{{dx}}\left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right){dx}\:=−\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=−\mathrm{2}\left[{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} =−\mathrm{2}\frac{\pi}{\mathrm{4}}\:=−\frac{\pi}{\mathrm{2}} \\ $$