Question Number 66792 by mathmax by abdo last updated on 19/Aug/19
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{cos}\left(\mathrm{2}\:{arctan}\left({x}\right)\right){dx} \\ $$
Commented by mathmax by abdo last updated on 20/Aug/19
![let I =∫_0 ^1 cos(2arctanx)dx we have cos(2arctanx) 2cos^2 (arctanx)−1 =2((1/(√(1+x^2 ))))^2 −1 =(2/(1+x^2 ))−1 ⇒ I =∫_0 ^1 ((2/(1+x^2 ))−1)dx =[2arctanx−x]_0 ^1 =2(π/4)−1 =(π/2)−1 .](Q66908.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{cos}\left(\mathrm{2}{arctanx}\right){dx}\:{we}\:{have}\:{cos}\left(\mathrm{2}{arctanx}\right) \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \left({arctanx}\right)−\mathrm{1}\:=\mathrm{2}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)^{\mathrm{2}} −\mathrm{1}\:=\frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}\right){dx}\:=\left[\mathrm{2}{arctanx}−{x}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{2}\frac{\pi}{\mathrm{4}}−\mathrm{1}\:=\frac{\pi}{\mathrm{2}}−\mathrm{1}\:. \\ $$