Question Number 43999 by maxmathsup by imad last updated on 19/Sep/18
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{x}}{\:\sqrt{\mathrm{1}−{cosx}}}{dx}. \\ $$
Commented by maxmathsup by imad last updated on 20/Sep/18
![let I = ∫_0 ^(π/2) (x/( (√(1−cosx))))dx ⇒I =∫_0 ^(π/2) (x/( (√(2sin^2 ((x/2))))))dx =(1/( (√2))) ∫_0 ^(π/2) (x/(sin((x/2))))dx =_((x/2)=t) (1/( (√2))) ∫_0 ^(π/4) ((2t)/(sin(t))) 2 dt =(4/( (√2))) ∫_0 ^(π/4) (t/(sint))dt changement tan((t/2))=u give I = (4/( (√2))) ∫_0 ^((√2)−1) ((2arctan(u))/((2u)/(1+u^2 ))) ((2du)/(1+u^2 )) =4(√( 2)) ∫_0 ^((√2)−1) ((arctan(u))/u) du let f(x)= ∫_0 ^((√2)−1) ((arctan(xu))/u) du ⇒ I =4(√2) f(1) we have f^′ (x) = ∫_0 ^((√2)−1) (du/(1+x^2 u^2 )) =_(xu =α) ∫_0 ^(x((√2)−1)) (1/(1+α^2 )) (dα/x) = (1/x) ∫_0 ^(x((√2)−1)) (dα/(1+α^2 )) =(1/x) [arctan(α)]_0 ^(x((√2)−1)) =((arctan{x((√2)−1)})/x) ⇒f(x) =∫_0 ^x ((arctan{t((√2)−1)})/t) dt +k k=f(0) =0 ⇒f(x) =∫_0 ^x ((arctan{t((√2)−1)})/t)dt ⇒ f(1) =∫_0 ^1 ((arctan{t((√2) −1)})/t)dtwe have arctan^′ (x) =(1/(1+x^2 )) =Σ_(n=0) ^∞ (−1)^n x^(2n) ⇒arctanx =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) x^(2n+1) ⇒ arctan{t((√2)−1)} =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) t^(2n+1) ((√2)−1)^(2n+1) ⇒ f(1) =Σ_(n=0) ^∞ ∫_0 ^1 (((−1)^n ((√2)−1)^(2n+1) )/(2n+1)) t^(2n) dt =Σ_(n=0) ^∞ (((−1)^n ((√2)−1)^(2n+1) )/((2n+1)^2 )) and I =4(√2)f(1)...](https://www.tinkutara.com/question/Q44033.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{x}}{\:\sqrt{\mathrm{1}−{cosx}}}{dx}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{x}}{\:\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}}{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{x}}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{dx}\:=_{\frac{{x}}{\mathrm{2}}={t}} \:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{\mathrm{2}{t}}{{sin}\left({t}\right)}\:\mathrm{2}\:{dt}\:=\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{t}}{{sint}}{dt} \\ $$$${changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{give}\: \\ $$$${I}\:\:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{\mathrm{2}{arctan}\left({u}\right)}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\mathrm{4}\sqrt{\:\mathrm{2}}\:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{{arctan}\left({u}\right)}{{u}}\:{du} \\ $$$${let}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{{arctan}\left({xu}\right)}{{u}}\:{du}\:\Rightarrow\:{I}\:=\mathrm{4}\sqrt{\mathrm{2}}\:{f}\left(\mathrm{1}\right) \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{{du}}{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }\:=_{{xu}\:=\alpha} \:\:\int_{\mathrm{0}} ^{{x}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:\frac{{d}\alpha}{{x}} \\ $$$$=\:\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{{x}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)} \:\:\frac{{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{x}}\:\left[{arctan}\left(\alpha\right)\right]_{\mathrm{0}} ^{{x}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)} \\ $$$$=\frac{{arctan}\left\{{x}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\right\}}{{x}}\:\Rightarrow{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{arctan}\left\{{t}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\right\}}{{t}}\:{dt}\:+{k} \\ $$$${k}={f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\:\frac{{arctan}\left\{{t}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\right\}}{{t}}{dt}\:\:\Rightarrow \\ $$$${f}\left(\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{arctan}\left\{{t}\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)\right\}}{{t}}{dtwe}\:{have} \\ $$$${arctan}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} \:\:\:\Rightarrow{arctanx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \Rightarrow \\ $$$${arctan}\left\{{t}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\right\}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{t}^{\mathrm{2}{n}+\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} \:\:\Rightarrow \\ $$$${f}\left(\mathrm{1}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:{t}^{\mathrm{2}{n}} {dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:{and}\:{I}\:=\mathrm{4}\sqrt{\mathrm{2}}{f}\left(\mathrm{1}\right)… \\ $$$$ \\ $$$$ \\ $$