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Question Number 117342 by bemath last updated on 11/Oct/20

(1)∫ (tan^(−1) (x))^2 dx = ? (2) ∫ tan^(−1) ((√x)) dx =?

$$\:\left(\mathrm{1}\right)\int\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} \:\mathrm{dx}\:=\:? \\ $$$$\left(\mathrm{2}\right)\:\int\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{x}}\right)\:\mathrm{dx}\:=? \\ $$

Answered by mathmax by abdo last updated on 11/Oct/20

I =∫ arctan^2 x dx by psrts I =x arctan^2 x−∫ x(((2arctanx))/(1+x^2 ))dx =x arctan^2 x−∫ ((2x)/(1+x^2 )) arctan(x)dx also by parts ∫ ((2x)/(1+x^2 )) arctan(x)dx =ln(1+x^2 )arctan(x)−∫ ((ln(1+x^2 ))/(1+x^2 ))dx ∫ ((ln(1+x^2 ))/(1+x^2 ))dx =_(x=tanθ) ∫ ((ln(1+tan^2 θ))/(1+tan^2 θ))(1+tan^2 θ)dθ =∫ ln((1/(cos^2 θ)))dθ =−2 ∫ ln(cosθ)dθ ⇒ I =x arctan^2 x−ln(1+x^2 )arctanx −2 ∫ ln(cosθ)dθ...

$$\mathrm{I}\:=\int\:\mathrm{arctan}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:\:\mathrm{by}\:\mathrm{psrts}\:\mathrm{I}\:=\mathrm{x}\:\mathrm{arctan}^{\mathrm{2}} \mathrm{x}−\int\:\mathrm{x}\frac{\left(\mathrm{2arctanx}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\mathrm{x}\:\mathrm{arctan}^{\mathrm{2}} \mathrm{x}−\int\:\:\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{arctan}\left(\mathrm{x}\right)\mathrm{dx}\:\:\mathrm{also}\:\mathrm{by}\:\mathrm{parts} \\ $$$$\int\:\:\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{arctan}\left(\mathrm{x}\right)\mathrm{dx}\:=\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{arctan}\left(\mathrm{x}\right)−\int\:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\int\:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=_{\mathrm{x}=\mathrm{tan}\theta} \:\:\int\:\:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)\mathrm{d}\theta \\ $$$$=\int\:\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta}\right)\mathrm{d}\theta\:=−\mathrm{2}\:\int\:\mathrm{ln}\left(\mathrm{cos}\theta\right)\mathrm{d}\theta\:\Rightarrow \\ $$$$\mathrm{I}\:=\mathrm{x}\:\mathrm{arctan}^{\mathrm{2}} \mathrm{x}−\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{arctanx}\:\:−\mathrm{2}\:\int\:\mathrm{ln}\left(\mathrm{cos}\theta\right)\mathrm{d}\theta... \\ $$

Commented by mathmax by abdo last updated on 11/Oct/20

thank you but what mean ∣...∣ because at C ∣re^(iθ) ∣ =r and ∣x+iy∣ =(√(x^2 +y^2 ))...!

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{but}\:\mathrm{what}\:\mathrm{mean}\:\mid...\mid\:\:\mathrm{because}\:\:\mathrm{at}\:\mathrm{C} \\ $$$$\mid\mathrm{re}^{\mathrm{i}\theta} \mid\:=\mathrm{r}\:\:\mathrm{and}\:\mid\mathrm{x}+\mathrm{iy}\mid\:=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }...! \\ $$

Commented by bemath last updated on 11/Oct/20

the last term ∫ ln (cos θ) dθ =?

$$\mathrm{the}\:\mathrm{last}\:\mathrm{term}\: \\ $$$$\int\:\mathrm{ln}\:\left(\mathrm{cos}\:\theta\right)\:\mathrm{d}\theta\:=? \\ $$

Commented by Lordose last updated on 11/Oct/20

check the pic i uploaded. it can also be expressed using clausen function of second order (Cl_2 )

$$\mathrm{check}\:\mathrm{the}\:\mathrm{pic}\:\mathrm{i}\:\mathrm{uploaded}. \\ $$$$\mathrm{it}\:\mathrm{can}\:\mathrm{also}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{using}\:\mathrm{clausen} \\ $$$$\mathrm{function}\:\mathrm{of}\:\mathrm{second}\:\mathrm{order}\:\left(\mathrm{Cl}_{\mathrm{2}} \right) \\ $$

Answered by bemath last updated on 11/Oct/20

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