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Question Number 42593 by aseerimad last updated on 28/Aug/18

Commented by maxmathsup by imad last updated on 28/Aug/18

we have I = ∫_0 ^(π/2)    ((cos^4 x +sin^4  x−sin^4 x)/(sin^4 x +cos^4 x))dx =(π/2) −∫_0 ^(π/2)   ((sin^4 x)/(sin^4 x +cos^4 x))dx but  changement x=(π/2)−t give ∫_0 ^(π/2)   ((sin^4 x)/(sin^4 x +cos^4 x))dx = ∫_0 ^(π/2)  ((cos^4 t)/(cos^4 x +sin^4 x))dt  =I ⇒ 2I =(π/2) ⇒ I =(π/4) .

$${we}\:{have}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{cos}^{\mathrm{4}} {x}\:+{sin}^{\mathrm{4}} \:{x}−{sin}^{\mathrm{4}} {x}}{{sin}^{\mathrm{4}} {x}\:+{cos}^{\mathrm{4}} {x}}{dx}\:=\frac{\pi}{\mathrm{2}}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sin}^{\mathrm{4}} {x}}{{sin}^{\mathrm{4}} {x}\:+{cos}^{\mathrm{4}} {x}}{dx}\:{but} \\ $$$${changement}\:{x}=\frac{\pi}{\mathrm{2}}−{t}\:{give}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sin}^{\mathrm{4}} {x}}{{sin}^{\mathrm{4}} {x}\:+{cos}^{\mathrm{4}} {x}}{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cos}^{\mathrm{4}} {t}}{{cos}^{\mathrm{4}} {x}\:+{sin}^{\mathrm{4}} {x}}{dt} \\ $$$$={I}\:\Rightarrow\:\mathrm{2}{I}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{4}}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Aug/18

I=∫_0 ^(Π/2) f(x)dx=∫_0 ^(Π/4) f((Π/2)−x)dx  I=∫_0 ^(Π/2) ((cos^4 x)/(sin^4 x+cos^4 x))dx=∫_0 ^(Π/4) ((cos^4 ((Π/2)−x))/(sin^4 ((Π/2)−x)+cos^4 ((Π/2)−x)))dx  I=∫_0 ^(Π/2) ((sin^4 x)/(cos^4 x+sin^4 x))dx  2I=∫_0 ^(Π/2) ((cos^4 x+sin^4 x)/(cos^4 x+sin^4 x))dx=∫_0 ^(Π/2)  dx  2I=∣x∣_0 ^(Π/2) =(Π/2)  I=(Π/4)

$${I}=\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} {f}\left(\frac{\Pi}{\mathrm{2}}−{x}\right){dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{4}} {x}}{{sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}}{dx}=\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} \frac{{cos}^{\mathrm{4}} \left(\frac{\Pi}{\mathrm{2}}−{x}\right)}{{sin}^{\mathrm{4}} \left(\frac{\Pi}{\mathrm{2}}−{x}\right)+{cos}^{\mathrm{4}} \left(\frac{\Pi}{\mathrm{2}}−{x}\right)}{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{4}} {x}}{{cos}^{\mathrm{4}} {x}+{sin}^{\mathrm{4}} {x}}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{4}} {x}+{sin}^{\mathrm{4}} {x}}{{cos}^{\mathrm{4}} {x}+{sin}^{\mathrm{4}} {x}}{dx}=\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:{dx} \\ $$$$\mathrm{2}{I}=\mid{x}\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} =\frac{\Pi}{\mathrm{2}} \\ $$$${I}=\frac{\Pi}{\mathrm{4}} \\ $$$$ \\ $$

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