Question Number 32480 by prof Abdo imad last updated on 25/Mar/18
$${find}\:\:\int_{\mathrm{0}} ^{\alpha} \:\sqrt{{tanx}}\:{dx}\:{with}\:\mathrm{0}<\alpha<\frac{\pi}{\mathrm{2}}\:. \\ $$
Commented by prof Abdo imad last updated on 14/Apr/18
![let put I = ∫_0 ^α (√(tanx)) dx .changement (√(tanx)) =t give x =arctan(t^2 ) and I = ∫_0 ^(√(tanα)) t .((2tdt)/(1+t^4 )) = ∫_0 ^(√(tanα)) ((2t^2 )/(1+t^4 )) dt = (1/2) ∫_0 ^(√(tanα)) (1/t) ((4t^3 )/(1+t^4 )) dt by parts u =(1/t) and v^′ = ((4t^3 )/(1+t^4 )) ⇒ I = [(1/t) ln(1+t^4 )]_0 ^(√(tanα)) − ∫_0 ^(√(tanα)) ((−1)/t^2 ) ln(1+t^4 ) dt = (1/( (√(tanα)))) ln( 1 +tan^2 α) + ∫_0 ^(√(tanα)) ((ln(1+t^4 ))/t^2 ) dt if 0< α< (π/4) ⇒ 0<(√(tanα))<1 so let developp ln(1+t^4 ) we have ln(1 +u)^′ =Σ (−1)^n u^n ln(1+u) =Σ _(n≥0) (((−1)^n u^(n+1) )/(n+1)) = Σ_(n=1) ^∞ (−1)^(n−1) (u^n /n) ln(1+x^4 ) = Σ_(n=1) ^∞ (−1)^(n−1) (x^(4n) /n) ⇒ ((ln(1+t^4 ))/t^2 ) = Σ_(n=1) ^∞ (−1)^(n−1) (t^(4n−2) /n) ⇒ ∫_0 ^(√(tanα)) ((ln(1+t^4 ))/t^2 ) dt = Σ_(n=1) ^∞ (((−1)^(n−1) )/n) ∫_0 ^(√(tanα)) t^(4n−2) dt = Σ_(n=1) ^∞ (((−1)^(n−1) )/n) (1/(4n−1)) (tanα)^(2n−1) let put S(x) = Σ_(n=1) ^∞ (((−1)^(n−1) )/n) (1/(4n−1)) x^(2n−1) let find S(x) (1/4) S(x) = Σ_(n=1) ^∞ (((−1)^(n−1) )/((4n−1)(4n))) x^(2n−1) = Σ_(n=1) ^∞ (−1)^(n−1) ( (1/(4n−1)) −(1/(4n))) x^(2n−1) =Σ_(n=1) ^∞ (((−1)^(n−1) )/(4n−1)) x^(2n−1) −(1/4) Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^(2n−1) ...be continued...](https://www.tinkutara.com/question/Q33260.png)
$${let}\:{put}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\alpha} \:\sqrt{{tanx}}\:{dx}\:\:.{changement}\:\sqrt{{tanx}}\:={t} \\ $$$${give}\:{x}\:={arctan}\left({t}^{\mathrm{2}} \right)\:\:{and}\: \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:\:\:{t}\:.\frac{\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:\:=\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\frac{\mathrm{1}}{{t}}\:\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}\:\:{by}\:{parts}\:{u}\:=\frac{\mathrm{1}}{{t}}\:{and} \\ $$$${v}^{'} \:=\:\:\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:\Rightarrow\:{I}\:=\:\left[\frac{\mathrm{1}}{{t}}\:{ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)\right]_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \\ $$$$−\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:\frac{−\mathrm{1}}{{t}^{\mathrm{2}} }\:{ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)\:{dt} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{{tan}\alpha}}\:{ln}\left(\:\mathrm{1}\:+{tan}^{\mathrm{2}} \alpha\right)\:+\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{{t}^{\mathrm{2}} }\:{dt} \\ $$$${if}\:\:\mathrm{0}<\:\alpha<\:\frac{\pi}{\mathrm{4}}\:\:\Rightarrow\:\mathrm{0}<\sqrt{{tan}\alpha}<\mathrm{1}\:\:{so}\:{let}\:{developp} \\ $$$${ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)\:\:{we}\:{have}\:\:{ln}\left(\mathrm{1}\:+{u}\right)^{'} \:=\Sigma\:\left(−\mathrm{1}\right)^{{n}} {u}^{{n}} \\ $$$${ln}\left(\mathrm{1}+{u}\right)\:=\Sigma\:_{{n}\geqslant\mathrm{0}} \frac{\left(−\mathrm{1}\right)^{{n}} {u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{u}^{{n}} }{{n}} \\ $$$${ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{x}^{\mathrm{4}{n}} }{{n}}\:\Rightarrow \\ $$$$\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{{t}^{\mathrm{2}} }\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{t}^{\mathrm{4}{n}−\mathrm{2}} }{{n}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{{t}^{\mathrm{2}} }\:{dt}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:{t}^{\mathrm{4}{n}−\mathrm{2}} {dt} \\ $$$$=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}\:\left({tan}\alpha\right)^{\mathrm{2}{n}−\mathrm{1}} \:{let}\:{put} \\ $$$${S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}\:{x}^{\mathrm{2}{n}−\mathrm{1}} \:{let}\:{find}\:{S}\left({x}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\:{S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{4}{n}−\mathrm{1}\right)\left(\mathrm{4}{n}\right)}\:{x}^{\mathrm{2}{n}−\mathrm{1}} \\ $$$$=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left(\:\:\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{4}{n}}\right)\:{x}^{\mathrm{2}{n}−\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{4}{n}−\mathrm{1}}\:{x}^{\mathrm{2}{n}−\mathrm{1}} \:\:\:−\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{\mathrm{2}{n}−\mathrm{1}} \\ $$$$…{be}\:{continued}… \\ $$