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Question Number 57235 by maxmathsup by imad last updated on 31/Mar/19
calculate ∫_0 ^(π/2)  ((cosx −sinx)/( (√(cos^8 x +sin^8 x)))) dx
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cosx}\:−{sinx}}{\:\sqrt{{cos}^{\mathrm{8}} {x}\:+{sin}^{\mathrm{8}} {x}}}\:{dx} \\ $$
Commented by maxmathsup by imad last updated on 01/Apr/19
changement x=(π/2)−t give I =−∫_0 ^(π/2)  ((cos((π/2)−t)−sin((π/2)−t))/( (√(cos^8 ((π/2)−t)+sin^8 ((π/2)(t)))))(−dt)  =∫_0 ^(π/2)    ((sint −cost)/( (√(sin^8 t +cos^8 t)))) dt =−I ⇒2I =0 ⇒I =0 .
$${changement}\:{x}=\frac{\pi}{\mathrm{2}}−{t}\:{give}\:{I}\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cos}\left(\frac{\pi}{\mathrm{2}}−{t}\right)−{sin}\left(\frac{\pi}{\mathrm{2}}−{t}\right)}{\:\sqrt{{cos}^{\mathrm{8}} \left(\frac{\pi}{\mathrm{2}}−{t}\right)+{sin}^{\mathrm{8}} \left(\frac{\pi}{\mathrm{2}}\left({t}\right)\right.}}\left(−{dt}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sint}\:−{cost}}{\:\sqrt{{sin}^{\mathrm{8}} {t}\:+{cos}^{\mathrm{8}} {t}}}\:{dt}\:=−{I}\:\Rightarrow\mathrm{2}{I}\:=\mathrm{0}\:\Rightarrow{I}\:=\mathrm{0}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19
I=∫_0 ^(π/2) ((cosx−sinx)/( (√(cos^8 x+sin^8 x))))dx  =∫_0 ^(π/2) ((sinx−cosx)/( (√(sin^8 x+cos^8 x))))dx [∫_0 ^a f(x)dx=∫_0 ^a f(a−x)dx]  2I=0→I=0
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cosx}−{sinx}}{\:\sqrt{{cos}^{\mathrm{8}} {x}+{sin}^{\mathrm{8}} {x}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sinx}−{cosx}}{\:\sqrt{{sin}^{\mathrm{8}} {x}+{cos}^{\mathrm{8}} {x}}}{dx}\:\left[\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{a}} {f}\left({a}−{x}\right){dx}\right] \\ $$$$\mathrm{2}{I}=\mathrm{0}\rightarrow{I}=\mathrm{0} \\ $$

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