Question Number 144244 by SOMEDAVONG last updated on 23/Jun/21
$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{5tan}^{\mathrm{4}} \mathrm{x}+\mathrm{3cot}^{\mathrm{4}} \mathrm{x}}{\mathrm{tan}^{\mathrm{4}} \mathrm{x}+\mathrm{cot}^{\mathrm{4}} \mathrm{x}}\mathrm{dx}=? \\ $$
Answered by Dwaipayan Shikari last updated on 23/Jun/21
$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{5}{tg}^{\mathrm{4}} {x}+{ct}^{\mathrm{4}} {x}}{{tg}^{\mathrm{4}} {x}+{ct}^{\mathrm{4}} {x}}{dx}\overset{\int_{\delta} ^{\zeta} {f}\left({x}\right){dx}=\int_{\delta} ^{\zeta} {f}\left(\zeta+\delta−{x}\right){dx}} {=}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{3}{tg}^{\mathrm{4}} {x}+\mathrm{5}{cot}^{\mathrm{4}} {x}}{{tg}^{\mathrm{4}} {x}+{ct}^{\mathrm{4}} {x}}={I} \\ $$$$\mathrm{2}{I}=\mathrm{8}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{1}{dx}\Rightarrow{I}=\mathrm{2}\pi \\ $$
Commented by SOMEDAVONG last updated on 24/Jun/21
$$\mathrm{Thanks}\:\mathrm{sir}! \\ $$