Integration Questions

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 byAr Brandon last updated on 01/Apr/20

$${I}\:\:{love}\:\:{this}.\:\:{Great}\:\:{idea}!!! \\$$ $$\\$$

Commented byM±th+et£s last updated on 01/Apr/20

$${god}\:{bless}\:{you} \\$$

Commented byTANMAY PANACEA. last updated on 01/Apr/20

$${thank}\:{you}\:{sir} \\$$