Question-43840

Question Number 43840 by peter frank last updated on 16/Sep/18

Commented by maxmathsup by imad last updated on 16/Sep/18

let I = ∫_0 ^(π/2) ((sin^4 x)/(cos^4 x +sin^4 x))dx and J = ∫_0 ^(π/2) ((cos^4 x)/(cos^4 x +sin^4 x))dx we have I +J = ∫_0 ^(π/2) dx =(π/2) and J −I =∫_0 ^(π/2) ((cos^4 x −sin^4 x)/(cos^4 x +sin^4 x))dx = ∫_0 ^(π/2) ((cos^2 x −sin^2 x)/((cos^2 x +sin^2 x)^2 −2cos^2 x sin^2 x))dx =∫_0 ^(π/2) ((cos(2x))/(1−(1/2)sin^2 (2x)))dx = ∫_0 ^(π/2) ((2cos(2x))/(2−(1−cos^2 (2x))))dx = ∫_0 ^(π/2) ((2cos(2x))/(1+cos^2 (2x)))dx =_(2x=t) ∫_0 ^π ((2cost)/(1+cos^2 t)) dt = _(tan((t/2))=u) ∫_0 ^∞ ((2 ((1−u^2 )/(1+u^2 )))/(1+(((1−u^2 )/(1+u^2 )))^2 )) ((2du)/(1+u^2 )) = 4 ∫_0 ^∞ ((1−u^2 )/((1+u^2 )^2 { (((1+u^2 )^2 +(1−u^2 )^2 )/((1+u^2 )^2 ))}))du = 4 ∫_0 ^∞ ((1−u^2 )/(u^4 +2u^2 +1 +u^4 −2u^2 +1))du =4 ∫_0 ^∞ ((1−u^2 )/(2(u^4 +1)))du = 2 ∫_0 ^∞ ((1−u^2 )/(1+u^4 )) du =2 ∫_0 ^∞ (du/(1+u^4 )) −2 ∫_0 ^∞ (u^2 /(1+u^4 )) du but ∫_0 ^∞ (du/(1+u^4 )) =(π/(2(√2))) changement u=α^(1/4) give ∫_0 ^∞ (α^(1/2) /(1+α)) (1/4) α^((1/4)−1) dα = (1/4) ∫_0 ^∞ (α^((3/4)−1) /(1+α)) dα =(1/4) (π/(sin(((3π)/4)))) =(π/(4 ((√2)/2))) =(π/(2(√2))) ⇒ J−I = (π/( (√2))) −(π/( (√2))) =0 ⇒I =J ⇒ 2I =(π/2) ⇒ I =(π/4) and J =(π/4) .

$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sin}^{\mathrm{4}} {x}}{{cos}^{\mathrm{4}} {x}\:+{sin}^{\mathrm{4}} {x}}{dx}\:\:{and}\:{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cos}^{\mathrm{4}} {x}}{{cos}^{\mathrm{4}} {x}\:+{sin}^{\mathrm{4}} {x}}{dx}\:{we}\:{have}\: \\ $$$${I}\:+{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{dx}\:=\frac{\pi}{\mathrm{2}}\:{and}\:\:{J}\:−{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cos}^{\mathrm{4}} {x}\:−{sin}^{\mathrm{4}} {x}}{{cos}^{\mathrm{4}} {x}\:+{sin}^{\mathrm{4}} {x}}{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cos}^{\mathrm{2}} {x}\:−{sin}^{\mathrm{2}} {x}}{\left({cos}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} \:−\mathrm{2}{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}}{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\mathrm{2}{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}−\left(\mathrm{1}−{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)}{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{2}{cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{dx} \\ $$$$=_{\mathrm{2}{x}={t}} \:\:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{\mathrm{2}{cost}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:{dt}\:=\:_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{2}\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} \left\{\:\frac{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\}}{du} \\ $$$$=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{{u}^{\mathrm{4}} \:+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}\:+{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}}{du}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{2}\left({u}^{\mathrm{4}} \:+\mathrm{1}\right)}{du} \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{4}} }\:{du}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{4}} }\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{4}} }\:{du}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{4}} }\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:\:{changement}\:{u}=\alpha^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\alpha^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+\alpha}\:\frac{\mathrm{1}}{\mathrm{4}}\:\alpha^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {d}\alpha\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\alpha^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+\alpha}\:{d}\alpha\:=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{4}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${J}−{I}\:=\:\frac{\pi}{\:\sqrt{\mathrm{2}}}\:−\frac{\pi}{\:\sqrt{\mathrm{2}}}\:=\mathrm{0}\:\Rightarrow{I}\:={J}\:\Rightarrow\:\mathrm{2}{I}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{4}}\:\:{and}\:{J}\:=\frac{\pi}{\mathrm{4}}\:. \\ $$

Answered by Joel578 last updated on 16/Sep/18

∫_a ^b f(x) dx = ∫_a ^b f(a + b − x) dx I = ∫_0 ^(π/2) ((sin^4 x)/(cos^4 x + sin^4 x)) dx ... (i) using formula above gives us: I = ∫_0 ^(π/2) ((sin^4 ((π/2) − x))/(cos^4 ((π/2) − x) + sin^4 ((π/2) − x))) dx = ∫_0 ^(π/2) ((cos^4 x)/(sin^4 x + cos^4 x)) dx ... (ii) (i) + (ii) 2I = ∫_0 ^(π/2) ((sin^4 x)/(cos^4 x + sin^4 x)) dx + ∫_0 ^(π/2) ((cos^4 x)/(sin^4 x + cos^4 x)) dx = ∫_0 ^(π/2) 1 dx = (π/2) I = (π/4)

$$\int_{{a}} ^{{b}} \:{f}\left({x}\right)\:{dx}\:=\:\int_{{a}} ^{{b}} \:{f}\left({a}\:+\:{b}\:−\:{x}\right)\:{dx} \\ $$$$ \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{sin}^{\mathrm{4}} \:{x}}{\mathrm{cos}^{\mathrm{4}} \:{x}\:+\:\mathrm{sin}^{\mathrm{4}} \:{x}}\:{dx}\:\:\:\:\:\:\:\:\:\:\:…\:\left({i}\right) \\ $$$$\mathrm{using}\:\mathrm{formula}\:\mathrm{above}\:\mathrm{gives}\:\mathrm{us}: \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{sin}^{\mathrm{4}} \:\left(\frac{\pi}{\mathrm{2}}\:−\:{x}\right)}{\mathrm{cos}^{\mathrm{4}} \:\left(\frac{\pi}{\mathrm{2}}\:−\:{x}\right)\:+\:\mathrm{sin}^{\mathrm{4}} \:\left(\frac{\pi}{\mathrm{2}}\:−\:{x}\right)}\:{dx} \\ $$$$\:\:\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cos}^{\mathrm{4}} \:{x}}{\mathrm{sin}^{\mathrm{4}} \:{x}\:+\:\mathrm{cos}^{\mathrm{4}} \:{x}}\:{dx}\:\:\:\:\:\:\:\:\:\:…\:\left({ii}\right) \\ $$$$ \\ $$$$\left({i}\right)\:+\:\left({ii}\right) \\ $$$$\mathrm{2}{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{sin}^{\mathrm{4}} \:{x}}{\mathrm{cos}^{\mathrm{4}} \:{x}\:+\:\mathrm{sin}^{\mathrm{4}} \:{x}}\:{dx}\:+\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cos}^{\mathrm{4}} \:{x}}{\mathrm{sin}^{\mathrm{4}} \:{x}\:+\:\mathrm{cos}^{\mathrm{4}} \:{x}}\:{dx} \\ $$$$\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{1}\:{dx} \\ $$$$\:\:\:\:\:\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$\:\:{I}\:=\:\frac{\pi}{\mathrm{4}} \\ $$

Commented by Joel578 last updated on 16/Sep/18