Question Number 27184 by abdo imad last updated on 02/Jan/18
$${calculate}\:{in}\:{terms}\:{of}\:{x}\:\:\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\:}} \frac{{dt}}{\mathrm{1}+{xsint}}\:. \\ $$
Commented by abdo imad last updated on 05/Jan/18
![let do the changement tan((t/2))= α⇔t=2arctanα f(x)= ∫_0 ^1 (((2dα)/(1+α^2 ))/(1+ ((2xα)/(1+α^2 )))) = ∫_0 ^1 ((2dα)/(1+α^2 +2xα))=∫_0 ^1 ((2dα)/(α^2 +2xα +1)) =∫_0 ^1 ((2dα)/((α+x)^2 +1−x^2 )) case 1 /x/<1 we do the ch. α+x=(√(1−x^2 )) t f(x)= ∫_(x/(√(1−x^2 ))) ^((x+1)/(√(1−x^2 ))) ((2(√(1−x^2 )) dt)/((1−x^2 )(1+t^2 ))) f(x)=(2/(√(1−x^2 ))) [arctan(t)]_(x/(√(1−x^2 ))) ^((√(1+x))/(√(1−x))) f(x)= (2/(√(1−x^2 ))) ( arctan(((√(1+x))/(√(1−x))) ) −arctan((x/(√(1−x^2 ))))) case2 if /x/>1 f(x)= (1/x) ∫_0 ^(π/2) (dt/(x^(−1) +sinx)) and we can use the same mthode due to /x^(−1) /<1.....](Q27377.png)
$${let}\:{do}\:{the}\:{changement}\:\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)=\:\alpha\Leftrightarrow{t}=\mathrm{2}{arctan}\alpha \\ $$$${f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }}{\mathrm{1}+\:\frac{\mathrm{2}{x}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }}\:\:=\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} \:+\mathrm{2}{x}\alpha}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{d}\alpha}{\alpha^{\mathrm{2}} +\mathrm{2}{x}\alpha\:+\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{d}\alpha}{\left(\alpha+{x}\right)^{\mathrm{2}} +\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${case}\:\mathrm{1}\:\:\:/{x}/<\mathrm{1}\:\:\:{we}\:{do}\:{the}\:{ch}.\:\:\alpha+{x}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{t} \\ $$$${f}\left({x}\right)=\:\int_{\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} ^{\frac{{x}+\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} \:\:\:\:\:\:\:\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}\:{dt}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:\left[{arctan}\left({t}\right)\right]_{\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} ^{\frac{\sqrt{\mathrm{1}+{x}}}{\sqrt{\mathrm{1}−{x}}}} \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\left(\:{arctan}\left(\frac{\sqrt{\mathrm{1}+{x}}}{\sqrt{\mathrm{1}−{x}}}\:\right)\:−{arctan}\left(\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)\right) \\ $$$${case}\mathrm{2}\:\:{if}\:/{x}/>\mathrm{1}\:\: \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{{x}^{−\mathrm{1}} +{sinx}}\:\:{and}\:{we}\:{can}\:{use}\:{the}\:{same}\:{mthode} \\ $$$${due}\:{to}\:/{x}^{−\mathrm{1}} /<\mathrm{1}..... \\ $$