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Question Number 57325 by turbo msup by abdo last updated on 02/Apr/19

calculate ∫_0 ^(π/2)  ((ln(1+sinx))/(sinx))dx

$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{ln}\left(\mathrm{1}+{sinx}\right)}{{sinx}}{dx} \\ $$

Commented by Abdo msup. last updated on 05/Apr/19

let f(t) =∫_0 ^(π/2)  ((ln(1+t sinx))/(sinx))dx  with 0≤t≤1  we have f^′ (t) =∫_0 ^(π/2)  ((sinx)/(sinx(1+tsinx)))dx  =∫_0 ^(π/2)    (dx/(1+tsinx)) =_(tan((x/2))=u)       ∫_0 ^1     ((2du)/((1+u^2 )(1+t ((2u)/(1+u^2 )))))  =∫_0 ^1     ((2du)/(1+u^2  +2tu)) =∫_0 ^1    ((2du)/(u^2  +2tu +1))  =∫_0 ^1    ((2du)/(u^2  +2tu +u^2  +1−u^2 )) =∫_0 ^1    ((2du)/((u+t)^2  +1−t^2 ))  =_(u+t =(√(1−t^2 ))α)      ∫_(t/(√(1−t2))) ^((1+t)/(√(1−t^2 )))     ((2(√(1−t^2 ))dα)/((1−t^2 )(1+α^2 )))  =(2/(√(1−t^2 ))) [ arctan(α)]_(t/(√(1−t^2 ))) ^((1+t)/(√(1−t^2 )))   =(2/(√(1−t^2 ))){ arctan(((1+t)/(√(1−t^2 ))))−arctan((t/(√(1−t^2 ))))} ⇒  f(t) = ∫  (2/(√(1−t^2 ))) arctan(((1+t)/(√(1−t^2 ))))dt  −∫  (2/(√(1−t^2 ))) arctan((t/(√(1−t^2 ))))dt +c ....be continued..=

$${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{ln}\left(\mathrm{1}+{t}\:{sinx}\right)}{{sinx}}{dx}\:\:{with}\:\mathrm{0}\leqslant{t}\leqslant\mathrm{1} \\ $$$${we}\:{have}\:{f}^{'} \left({t}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sinx}}{{sinx}\left(\mathrm{1}+{tsinx}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+{tsinx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}} \:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}\:\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+\mathrm{2}{tu}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}{tu}\:+\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}{tu}\:+{u}^{\mathrm{2}} \:+\mathrm{1}−{u}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{du}}{\left({u}+{t}\right)^{\mathrm{2}} \:+\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$=_{{u}+{t}\:=\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\alpha} \:\:\:\:\:\int_{\frac{{t}}{\sqrt{\mathrm{1}−{t}\mathrm{2}}}} ^{\frac{\mathrm{1}+{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}} \:\:\:\:\frac{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{d}\alpha}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\left[\:{arctan}\left(\alpha\right)\right]_{\frac{{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}} ^{\frac{\mathrm{1}+{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\left\{\:{arctan}\left(\frac{\mathrm{1}+{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\right)−{arctan}\left(\frac{{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\right)\right\}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\:\int\:\:\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{arctan}\left(\frac{\mathrm{1}+{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\right){dt} \\ $$$$−\int\:\:\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{arctan}\left(\frac{{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\right){dt}\:+{c}\:....{be}\:{continued}..= \\ $$

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