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Question Number 69133 by Henri Boucatchou last updated on 20/Sep/19

∫(x − tanx)^2 dx = ?

$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int\left(\boldsymbol{\mathrm{x}}\:−\:\boldsymbol{\mathrm{tanx}}\right)^{\mathrm{2}} \:\boldsymbol{\mathrm{dx}}\:\:=\:? \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 20/Sep/19

let A =∫(x−tanx)^2 dx ⇒A =∫(x^2 −2x tanx +tan^2 x)dx =∫ x^2 dx −2 ∫ xtanxdx +∫ tan^2 xdx we have ∫ x^2 dx =(x^3 /3) +c_1 ∫ tan^2 x =∫ (1+tan^2 x−1)dx =tanx −x c_2 ∫ x tanx dx =_(tanx=t) ∫ t arctant (dt/(1+t^2 )) =∫ (t/(1+t^2 )) arctan(t)dt =(1/2)ln(1+t^2 )arctant −(1/2)∫ ((ln(1+t^2 ))/(1+t^2 ))dt rest to find ∫ ((ln(1+t^2 ))/(1+t^2 ))dt let f(x)=∫ ((ln(1+xt^2 ))/(1+t^2 ))dt ⇒ f^′ (x) = ∫ (t^2 /((1+xt^2 )(1+t^2 )))dt =(1/(x ))∫ ((1+xt^2 −1)/((1+xt^2 )(1+t^2 )))dt =(1/x) arctan(t)−(1/x) ∫ (dt/((xt^2 +1)(t^2 +1))) rest to decompose F(t) =(1/((xt^2 +1)(t^(2 ) +1))) be continued....

$${let}\:{A}\:=\int\left({x}−{tanx}\right)^{\mathrm{2}} {dx}\:\Rightarrow{A}\:=\int\left({x}^{\mathrm{2}} −\mathrm{2}{x}\:{tanx}\:+{tan}^{\mathrm{2}} {x}\right){dx} \\ $$$$=\int\:{x}^{\mathrm{2}} {dx}\:−\mathrm{2}\:\int\:{xtanxdx}\:+\int\:{tan}^{\mathrm{2}} {xdx}\:\:{we}\:{have} \\ $$$$\int\:{x}^{\mathrm{2}} {dx}\:=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+{c}_{\mathrm{1}} \\ $$$$\int\:{tan}^{\mathrm{2}} {x}\:=\int\:\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}−\mathrm{1}\right){dx}\:={tanx}\:−{x}\:{c}_{\mathrm{2}} \\ $$$$\int\:\:{x}\:{tanx}\:{dx}\:=_{{tanx}={t}} \:\:\int\:{t}\:{arctant}\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int\:\:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{arctan}\left({t}\right){dt}\:\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){arctant}\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${rest}\:{to}\:{find}\:\int\:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{let}\:{f}\left({x}\right)=\int\:\frac{{ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\:\int\:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:=\frac{\mathrm{1}}{{x}\:}\int\:\:\frac{\mathrm{1}+{xt}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\frac{\mathrm{1}}{{x}}\:{arctan}\left({t}\right)−\frac{\mathrm{1}}{{x}}\:\int\:\:\:\frac{{dt}}{\left({xt}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\:\:\:{rest}\:{to}\:{decompose} \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{1}}{\left({xt}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}\:} +\mathrm{1}\right)}\:\:\:{be}\:{continued}.... \\ $$

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