Question Number 86995 by M±th+et£s last updated on 01/Apr/20 | ||
$$\int_{\mathrm{0}} ^{\pi} \frac{{a}^{{n}} {sin}^{\mathrm{2}} \left({x}\right)+{b}^{{n}} {cos}^{\mathrm{2}} \left({x}\right)}{{a}^{\mathrm{2}{n}} {sin}^{\mathrm{2}} \left({x}\right)+{b}^{\mathrm{2}{n}} {cos}^{\mathrm{2}} \left({x}\right)}{dx}\:;\:{a}>{b} \\ $$ | ||
Answered by TANMAY PANACEA. last updated on 01/Apr/20 | ||
$$\int_{\mathrm{0}} ^{\pi} \frac{{a}^{{n}} \left(\mathrm{1}−{cos}\mathrm{2}{x}\right)+{b}^{{n}} \left(\mathrm{1}+{cos}\mathrm{2}{x}\right)}{{a}^{\mathrm{2}{n}} \left(\mathrm{1}−{cos}\mathrm{2}{x}\right)+{b}^{\mathrm{2}{n}} \left(\mathrm{1}+{cos}\mathrm{2}{x}\right)}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{a}^{{n}} +{b}^{{n}} −\left({a}^{{n}} −{b}^{{n}} \right){cos}\mathrm{2}{x}}{{a}^{\mathrm{2}{n}} +{b}^{\mathrm{2}{n}} −\left({a}^{\mathrm{2}{n}} −{b}^{\mathrm{2}{n}} \right){cos}\mathrm{2}{x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{A}−{Bcos}\mathrm{2}{x}}{{C}−{Dcos}\mathrm{2}{x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{A}}{{C}−{Dcos}\mathrm{2}{x}}{dx}+\frac{{B}}{{D}}\int_{\mathrm{0}} ^{\pi} \frac{{C}−{Dcos}\mathrm{2}{x}−{C}}{{C}−{Dcos}\mathrm{2}{x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{A}}{{C}−{Dcos}\mathrm{2}{x}}+\frac{{B}}{{D}}\int_{\mathrm{0}} ^{\pi} {dx}−\frac{{BC}}{{D}}\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{{C}−{Dcos}\mathrm{2}{x}} \\ $$$$\left({A}−\frac{{BC}}{{D}}\right)\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} {xdx}}{{C}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)−{D}\left(\mathrm{1}−{tan}^{\mathrm{2}} {x}\right)}+\frac{{B}}{{D}}\int_{\mathrm{0}} ^{\pi} {dx} \\ $$$$\left({A}−\frac{{BC}}{{D}}\right)\int_{\mathrm{0}} ^{\pi} \frac{{d}\left({tanx}\right)}{\left({C}−{D}\right)+\left({C}+{D}\right){tan}^{\mathrm{2}} {x}}+\frac{{B}}{{D}}\int_{\mathrm{0}} ^{\pi} {dx} \\ $$$$\frac{{AD}−{BC}}{{D}\left({C}+{D}\right)}\int_{\mathrm{0}} ^{\pi} \frac{{d}\left({tanx}\right)}{\frac{{C}−{D}}{{C}+{D}}+{tan}^{\mathrm{2}} {x}}+\frac{{B}}{{D}}\int_{\mathrm{0}} ^{\pi} {dx} \\ $$$$\frac{{AD}−{BC}}{{D}\left({C}+{D}\right)}×\frac{\mathrm{1}}{\sqrt{\frac{{C}−{D}}{{C}+{D}}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{tanx}}{\sqrt{\frac{{C}−{D}}{{C}+{D}}}}\right)\mid_{\mathrm{0}} ^{\pi} +\frac{{B}}{{D}}\pi \\ $$$$\frac{{B}}{{D}}\pi \\ $$$$=\frac{\left({a}^{{n}} −{b}^{{n}} \right)}{{a}^{\mathrm{2}{n}} −{b}^{\mathrm{2}{n}} }×\pi \\ $$$$=\frac{\pi}{{a}^{{n}} +{b}^{{n}} } \\ $$$${pls}\:{check}... \\ $$$$ \\ $$ | ||
Commented by Ar Brandon last updated on 01/Apr/20 | ||
$${I}\:\:{love}\:\:{this}.\:\:{Great}\:\:{idea}!!! \\ $$$$ \\ $$ | ||
Commented by M±th+et£s last updated on 01/Apr/20 | ||
$${god}\:{bless}\:{you} \\ $$ | ||
Commented by TANMAY PANACEA. last updated on 01/Apr/20 | ||
$${thank}\:{you}\:{sir} \\ $$ | ||