Question Number 186771 by mnjuly1970 last updated on 10/Feb/23
$$ \\ $$$$\:\:\:\:\mathrm{Q}\::\:\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{integral}.\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\:\frac{\:\pi}{\:\mathrm{2}}} \:\frac{\:\:\mathrm{1}}{\:\mathrm{1}\:+\:\mathrm{sin}^{\:\mathrm{4}} \:\left(\:{x}\:\right)\:+\:\mathrm{cos}^{\:\mathrm{4}} \:\left(\:{x}\:\right)\:}\:\mathrm{d}{x}\:=\:\:?\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 10/Feb/23
![I=∫_0 ^(π/2) (dx/(1+sin^4 x+cos^4 x))=∫_0 ^(π/2) ((sec^4 x)/(sec^4 x+tan^4 x+1))dx =∫_0 ^(π/2) ((tan^2 x+1)/((tan^2 x+1)^2 +tan^4 x+1))d(tanx) =∫_0 ^∞ ((t^2 +1)/(2t^4 +2t^2 +2))dt=(1/2)∫_0 ^∞ ((t^2 +1)/(t^4 +t^2 +1))dt =(1/2)∫_0 ^∞ ((1+(1/t^2 ))/(t^2 +1+(1/t^2 )))dt=(1/2)∫_0 ^∞ ((1+(1/t^2 ))/((t−(1/t))^2 +3))dt =(1/2)∫_(−∞) ^∞ (du/(u^2 +3))=(1/(2(√3)))[arctan((u/( (√3))))]_(−∞) ^∞ =(1/(2(√3)))((π/2)−−(π/2))=(1/(2(√3)))((π/2)+(π/2))=(π/(2(√3)))](Q186779.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{4}} {x}+\mathrm{cos}^{\mathrm{4}} {x}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}^{\mathrm{4}} {x}}{\mathrm{sec}^{\mathrm{4}} {x}+\mathrm{tan}^{\mathrm{4}} {x}+\mathrm{1}}{dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{1}}{\left(\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{tan}^{\mathrm{4}} {x}+\mathrm{1}}{d}\left(\mathrm{tan}{x}\right) \\ $$$$\:\:=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{3}}{dt} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{du}}{{u}^{\mathrm{2}} +\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left[\mathrm{arctan}\left(\frac{{u}}{\:\sqrt{\mathrm{3}}}\right)\right]_{−\infty} ^{\infty} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{\pi}{\mathrm{2}}−−\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$
Commented by mnjuly1970 last updated on 11/Feb/23
$${thanks}\:{slot}\:{sir} \\ $$
Answered by integralmagic last updated on 10/Feb/23
$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{arctan}\left(\sqrt{\mathrm{3}}/\mathrm{2}{tan}\mathrm{2}{x}\right)\mid_{\mathrm{0}} ^{\pi/\mathrm{2}} =\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$