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Question Number 32480 by prof Abdo imad last updated on 25/Mar/18

find  ∫_0 ^α  (√(tanx)) dx with 0<α<(π/2) .

$${find}\:\:\int_{\mathrm{0}} ^{\alpha} \:\sqrt{{tanx}}\:{dx}\:{with}\:\mathrm{0}<\alpha<\frac{\pi}{\mathrm{2}}\:. \\ $$

Commented byprof 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...

$${let}\:{put}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\alpha} \:\sqrt{{tanx}}\:{dx}\:\:.{changement}\:\sqrt{{tanx}}\:={t} \\ $$ $${give}\:{x}\:={arctan}\left({t}^{\mathrm{2}} \right)\:\:{and}\: \\ $$ $${I}\:=\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:\:\:{t}\:.\frac{\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:\:=\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt} \\ $$ $$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\frac{\mathrm{1}}{{t}}\:\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}\:\:{by}\:{parts}\:{u}\:=\frac{\mathrm{1}}{{t}}\:{and} \\ $$ $${v}^{'} \:=\:\:\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:\Rightarrow\:{I}\:=\:\left[\frac{\mathrm{1}}{{t}}\:{ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)\right]_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \\ $$ $$−\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:\frac{−\mathrm{1}}{{t}^{\mathrm{2}} }\:{ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)\:{dt} \\ $$ $$=\:\frac{\mathrm{1}}{\sqrt{{tan}\alpha}}\:{ln}\left(\:\mathrm{1}\:+{tan}^{\mathrm{2}} \alpha\right)\:+\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{{t}^{\mathrm{2}} }\:{dt} \\ $$ $${if}\:\:\mathrm{0}<\:\alpha<\:\frac{\pi}{\mathrm{4}}\:\:\Rightarrow\:\mathrm{0}<\sqrt{{tan}\alpha}<\mathrm{1}\:\:{so}\:{let}\:{developp} \\ $$ $${ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)\:\:{we}\:{have}\:\:{ln}\left(\mathrm{1}\:+{u}\right)^{'} \:=\Sigma\:\left(−\mathrm{1}\right)^{{n}} {u}^{{n}} \\ $$ $${ln}\left(\mathrm{1}+{u}\right)\:=\Sigma\:_{{n}\geqslant\mathrm{0}} \frac{\left(−\mathrm{1}\right)^{{n}} {u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{u}^{{n}} }{{n}} \\ $$ $${ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{x}^{\mathrm{4}{n}} }{{n}}\:\Rightarrow \\ $$ $$\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{{t}^{\mathrm{2}} }\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{t}^{\mathrm{4}{n}−\mathrm{2}} }{{n}}\:\Rightarrow \\ $$ $$\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{{t}^{\mathrm{2}} }\:{dt}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\int_{\mathrm{0}} ^{\sqrt{{tan}\alpha}} \:\:{t}^{\mathrm{4}{n}−\mathrm{2}} {dt} \\ $$ $$=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}\:\left({tan}\alpha\right)^{\mathrm{2}{n}−\mathrm{1}} \:{let}\:{put} \\ $$ $${S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}\:{x}^{\mathrm{2}{n}−\mathrm{1}} \:{let}\:{find}\:{S}\left({x}\right) \\ $$ $$\frac{\mathrm{1}}{\mathrm{4}}\:{S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{4}{n}−\mathrm{1}\right)\left(\mathrm{4}{n}\right)}\:{x}^{\mathrm{2}{n}−\mathrm{1}} \\ $$ $$=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left(\:\:\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{4}{n}}\right)\:{x}^{\mathrm{2}{n}−\mathrm{1}} \\ $$ $$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{4}{n}−\mathrm{1}}\:{x}^{\mathrm{2}{n}−\mathrm{1}} \:\:\:−\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{\mathrm{2}{n}−\mathrm{1}} \\ $$ $$...{be}\:{continued}... \\ $$

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