Question Number 57235 by maxmathsup by imad last updated on 31/Mar/19
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cosx}\:−{sinx}}{\sqrt{{cos}^{\mathrm{8}} {x}\:+{sin}^{\mathrm{8}} {x}}}\:{dx} \\ $$
Commented by maxmathsup by imad last updated on 01/Apr/19
$${changement}\:{x}=\frac{\pi}{\mathrm{2}}−{t}\:{give}\:{I}\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cos}\left(\frac{\pi}{\mathrm{2}}−{t}\right)−{sin}\left(\frac{\pi}{\mathrm{2}}−{t}\right)}{\sqrt{{cos}^{\mathrm{8}} \left(\frac{\pi}{\mathrm{2}}−{t}\right)+{sin}^{\mathrm{8}} \left(\frac{\pi}{\mathrm{2}}\left({t}\right)\right.}}\left(−{dt}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sint}\:−{cost}}{\sqrt{{sin}^{\mathrm{8}} {t}\:+{cos}^{\mathrm{8}} {t}}}\:{dt}\:=−{I}\:\Rightarrow\mathrm{2}{I}\:=\mathrm{0}\:\Rightarrow{I}\:=\mathrm{0}\:. \\ $$
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](Q57240.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cosx}−{sinx}}{\sqrt{{cos}^{\mathrm{8}} {x}+{sin}^{\mathrm{8}} {x}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sinx}−{cosx}}{\sqrt{{sin}^{\mathrm{8}} {x}+{cos}^{\mathrm{8}} {x}}}{dx}\:\left[\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{a}} {f}\left({a}−{x}\right){dx}\right] \\ $$$$\mathrm{2}{I}=\mathrm{0}\rightarrow{I}=\mathrm{0} \\ $$