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Question Number 87534 by mathmax by abdo last updated on 04/Apr/20

calculate ∫_0 ^(π/4) ((arctan(sinx))/(sinx))dx

$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{arctan}\left({sinx}\right)}{{sinx}}{dx} \\ $$

Commented by mathmax by abdo last updated on 06/Apr/20

let f(a) =∫_0 ^(π/4) ((arctan(asinx))/(sinx))dx with a>0 f^′ (a) =∫_0 ^(π/4) ((sinx)/((1+a^2 sin^2 x)sinx))dx =∫_0 ^(π/4) (dx/(1+a^2 ×((1−cos(2x))/2))) =2∫_0 ^(π/4) (dx/(2+a^2 −a^2 cos(2x))) =_(2x=t) 2 ∫_0 ^(π/2) (dt/(2(2+a^2 −a^2 cost))) =_(tan((t/2))=u) ∫_0 ^1 ((2du)/((1+u^2 )(2+a^2 −a^2 ((1−u^2 )/(1+u^2 ))))) =∫_0 ^1 ((2du)/(2+a^2 +(2+a^2 )u^2 −a^2 +a^2 u^2 )) =∫_0 ^1 ((2du)/(2 +(2+2a^2 )u^2 )) =∫_0 ^1 (du/(1+(1+a^2 )u^2 )) =_((√(1+a^2 ))u=z) ∫_0 ^(√(1+a^2 )) (dz/((√(1+a^2 ))(1+z^2 ))) =(1/(√(1+a^2 ))) arctan((√(1+a^2 ))) ⇒f(a) =∫_0 ^a ((arctan((√(1+α^2 ))))/(√(1+α^2 ))) dα +c c=f(0)=0 ⇒f(a) =∫_0 ^a ((arctan((√(1+α^2 ))))/(√(1+α^2 ))) dα and ∫_0 ^(π/4) ((arctan(sinx))/(sinx))dx =∫_0 ^1 ((arctan((√(1+x^2 ))))/(√(1+x^2 )))dx....be continued...

$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{arctan}\left({asinx}\right)}{{sinx}}{dx}\:{with}\:{a}>\mathrm{0} \\ $$$${f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{sinx}}{\left(\mathrm{1}+{a}^{\mathrm{2}} \:{sin}^{\mathrm{2}} {x}\right){sinx}}{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{dx}}{\mathrm{1}+{a}^{\mathrm{2}} ×\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{dx}}{\mathrm{2}+{a}^{\mathrm{2}} −{a}^{\mathrm{2}} {cos}\left(\mathrm{2}{x}\right)}\:=_{\mathrm{2}{x}={t}} \:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{\mathrm{2}\left(\mathrm{2}+{a}^{\mathrm{2}} −{a}^{\mathrm{2}} {cost}\right)} \\ $$$$=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}+{a}^{\mathrm{2}} −{a}^{\mathrm{2}} \frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{du}}{\mathrm{2}+{a}^{\mathrm{2}} \:+\left(\mathrm{2}+{a}^{\mathrm{2}} \right){u}^{\mathrm{2}} −{a}^{\mathrm{2}} \:+{a}^{\mathrm{2}} {u}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{du}}{\mathrm{2}\:+\left(\mathrm{2}+\mathrm{2}{a}^{\mathrm{2}} \right){u}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{du}}{\mathrm{1}+\left(\mathrm{1}+{a}^{\mathrm{2}} \right){u}^{\mathrm{2}} }\:=_{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }{u}={z}} \:\:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }} \:\:\:\frac{{dz}}{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\left(\mathrm{1}+{z}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)\:\Rightarrow{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{{a}} \:\frac{{arctan}\left(\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }\right)}{\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }}\:{d}\alpha\:+{c} \\ $$$${c}={f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{{a}} \:\frac{{arctan}\left(\sqrt{\left.\mathrm{1}+\alpha^{\mathrm{2}} \right)}\right.}{\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }}\:{d}\alpha \\ $$$${and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{arctan}\left({sinx}\right)}{{sinx}}{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}....{be}\:{continued}... \\ $$

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