Question Number 87052 by lémùst last updated on 02/Apr/20
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{sin}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}+{xsin}^{\mathrm{2}} \left({t}\right)}{dt} \\ $$
Commented by mathmax by abdo last updated on 02/Apr/20
![f(x)=∫_0 ^(π/2) ((sin^2 t)/(1+x sin^2 t))dt=(1/x)∫_0 ^(π/2) ((xsin^2 t +1−1)/(1+xsin^2 t))dt =(π/(2x))−(1/x) ∫_0 ^(π/(2 )) (dt/(1+xsin^2 t)) [we have ∫_0 ^(π/2) (dt/(1+xsin^2 t)) =∫_0 ^(π/2) (dt/(1+x×((1−cos(2t))/2))) =_(2x=u) ∫_0 ^π (du/(2(1+x×((1−cos(2u))/2)))) = ∫_0 ^π (du/(2+x−xcosu)) =_(tan((u/2))=z) ∫_0 ^∞ ((2dz)/((1+z^2 )(2+x −x((1−z^2 )/(1+z^2 ))))) =∫_0 ^∞ ((2dz)/(2+x +(2+x)z^2 −x+xz^2 )) =∫_0 ^∞ ((2dz)/(2+(2+2x)z^2 )) =∫_0 ^∞ (dz/(1+(1+x)z^2 )) case 1 1+x>0 we do changement u=(√(1+x))z ⇒ ∫_0 ^∞ (dz/(1+(1+x)z^2 )) =∫_0 ^∞ (du/((√(1+x))(1+u^2 ))) =(1/(√(1+x))) ×(π/2) ⇒ f(x)=(π/(2x))−(π/(2x(√(1+x)))) (x>−1 and x≠0) case 2 1+x<0 ⇒∫_0 ^∞ (dz/(1+(1+x)z^2 )) =∫_0 ^∞ (dz/(1−(−(1+x))z^2 )) =∫_0 ^∞ (dz/((1−(√(−1−x))z)(1+(√(−1−x))z))) =(1/2)∫_0 ^∞ ((1/(1−(√(−1−x))z))+(1/(1+(√(−1−x))z))) =(1/2)∫_0 ^∞ ((1/(−αz +1)) +(1/(αz +1)))dz (α=(√(−1−x))) =(1/2)[(1/α)ln∣αz +1∣−(1/α)ln∣αz−1∣]_0 ^(+∞) (1/(2α))[ln∣((αz +1)/(αz−1))∣]_0 ^(+∞) =0 so f(x)=(π/(2x))−(π/(2x(√(1+x)))) if x>−1 and x≠−1 f(x)=0 if x<−1](Q87058.png)
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sin}^{\mathrm{2}} {t}}{\mathrm{1}+{x}\:{sin}^{\mathrm{2}} {t}}{dt}=\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{xsin}^{\mathrm{2}} {t}\:+\mathrm{1}−\mathrm{1}}{\mathrm{1}+{xsin}^{\mathrm{2}} {t}}{dt} \\ $$$$=\frac{\pi}{\mathrm{2}{x}}−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\:}} \:\frac{{dt}}{\mathrm{1}+{xsin}^{\mathrm{2}} {t}}\:\left[{we}\:{have}\right. \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{\mathrm{1}+{xsin}^{\mathrm{2}} {t}}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{\mathrm{1}+{x}×\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}} \\ $$$$=_{\mathrm{2}{x}={u}} \:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{du}}{\mathrm{2}\left(\mathrm{1}+{x}×\frac{\mathrm{1}−{cos}\left(\mathrm{2}{u}\right)}{\mathrm{2}}\right)}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{du}}{\mathrm{2}+{x}−{xcosu}} \\ $$$$=_{{tan}\left(\frac{{u}}{\mathrm{2}}\right)={z}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dz}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)\left(\mathrm{2}+{x}\:−{x}\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{1}+{z}^{\mathrm{2}} }\right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dz}}{\mathrm{2}+{x}\:+\left(\mathrm{2}+{x}\right){z}^{\mathrm{2}} −{x}+{xz}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dz}}{\mathrm{2}+\left(\mathrm{2}+\mathrm{2}{x}\right){z}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dz}}{\mathrm{1}+\left(\mathrm{1}+{x}\right){z}^{\mathrm{2}} }\: \\ $$$${case}\:\mathrm{1}\:\:\:\mathrm{1}+{x}>\mathrm{0}\:\:{we}\:{do}\:{changement}\:{u}=\sqrt{\mathrm{1}+{x}}{z}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dz}}{\mathrm{1}+\left(\mathrm{1}+{x}\right){z}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\sqrt{\mathrm{1}+{x}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}}}\:×\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\pi}{\mathrm{2}{x}}−\frac{\pi}{\mathrm{2}{x}\sqrt{\mathrm{1}+{x}}}\:\:\left({x}>−\mathrm{1}\:{and}\:{x}\neq\mathrm{0}\right) \\ $$$${case}\:\mathrm{2}\:\:\mathrm{1}+{x}<\mathrm{0}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dz}}{\mathrm{1}+\left(\mathrm{1}+{x}\right){z}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dz}}{\mathrm{1}−\left(−\left(\mathrm{1}+{x}\right)\right){z}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dz}}{\left(\mathrm{1}−\sqrt{−\mathrm{1}−{x}}{z}\right)\left(\mathrm{1}+\sqrt{−\mathrm{1}−{x}}{z}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\left(\frac{\mathrm{1}}{\mathrm{1}−\sqrt{−\mathrm{1}−{x}}{z}}+\frac{\mathrm{1}}{\mathrm{1}+\sqrt{−\mathrm{1}−{x}}{z}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{−\alpha{z}\:+\mathrm{1}}\:+\frac{\mathrm{1}}{\alpha{z}\:+\mathrm{1}}\right){dz}\:\:\:\left(\alpha=\sqrt{−\mathrm{1}−{x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\alpha}{ln}\mid\alpha{z}\:+\mathrm{1}\mid−\frac{\mathrm{1}}{\alpha}{ln}\mid\alpha{z}−\mathrm{1}\mid\right]_{\mathrm{0}} ^{+\infty} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\alpha}\left[{ln}\mid\frac{\alpha{z}\:+\mathrm{1}}{\alpha{z}−\mathrm{1}}\mid\right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{0}\:\:{so} \\ $$$${f}\left({x}\right)=\frac{\pi}{\mathrm{2}{x}}−\frac{\pi}{\mathrm{2}{x}\sqrt{\mathrm{1}+{x}}}\:{if}\:{x}>−\mathrm{1}\:{and}\:{x}\neq−\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{0}\:{if}\:{x}<−\mathrm{1} \\ $$
Commented by lémùst last updated on 02/Apr/20
$${merci}\:{beaucoup}\: \\ $$$${mais}\:{je}\:{pense}\:{que}\:{f}\left({x}\right)=\frac{\pi}{\mathrm{2}{x}}\:{si}\:{x}<−\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 02/Apr/20
$${oui}\:\:{tu}\:{a}\:{raison}\:{j}\:{ai}\:{oublie}\:\frac{\pi}{\mathrm{2}{x}}\:{merci}\:. \\ $$
Commented by Ar Brandon last updated on 03/Apr/20
$${genial}!! \\ $$