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Question Number 86995 by M±th+et£s last updated on 01/Apr/20

∫_0 ^π ((a^n sin^2 (x)+b^n cos^2 (x))/(a^(2n) sin^2 (x)+b^(2n) cos^2 (x)))dx ; a>b

$$\int_{\mathrm{0}} ^{\pi} \frac{{a}^{{n}} {sin}^{\mathrm{2}} \left({x}\right)+{b}^{{n}} {cos}^{\mathrm{2}} \left({x}\right)}{{a}^{\mathrm{2}{n}} {sin}^{\mathrm{2}} \left({x}\right)+{b}^{\mathrm{2}{n}} {cos}^{\mathrm{2}} \left({x}\right)}{dx}\:;\:{a}>{b} \\ $$

Answered by TANMAY PANACEA. last updated on 01/Apr/20

∫_0 ^π ((a^n (1−cos2x)+b^n (1+cos2x))/(a^(2n) (1−cos2x)+b^(2n) (1+cos2x)))dx ∫_0 ^π ((a^n +b^n −(a^n −b^n )cos2x)/(a^(2n) +b^(2n) −(a^(2n) −b^(2n) )cos2x))dx ∫_0 ^π ((A−Bcos2x)/(C−Dcos2x))dx ∫_0 ^π (A/(C−Dcos2x))dx+(B/D)∫_0 ^π ((C−Dcos2x−C)/(C−Dcos2x))dx ∫_0 ^π (A/(C−Dcos2x))+(B/D)∫_0 ^π dx−((BC)/D)∫_0 ^π (dx/(C−Dcos2x)) (A−((BC)/D))∫_0 ^π ((sec^2 xdx)/(C(1+tan^2 x)−D(1−tan^2 x)))+(B/D)∫_0 ^π dx (A−((BC)/D))∫_0 ^π ((d(tanx))/((C−D)+(C+D)tan^2 x))+(B/D)∫_0 ^π dx ((AD−BC)/(D(C+D)))∫_0 ^π ((d(tanx))/(((C−D)/(C+D))+tan^2 x))+(B/D)∫_0 ^π dx ((AD−BC)/(D(C+D)))×(1/(√((C−D)/(C+D))))∣tan^(−1) (((tanx)/(√((C−D)/(C+D)))))∣_0 ^π +(B/D)π (B/D)π =(((a^n −b^n ))/(a^(2n) −b^(2n) ))×π =(π/(a^n +b^n )) pls check...

$$\int_{\mathrm{0}} ^{\pi} \frac{{a}^{{n}} \left(\mathrm{1}−{cos}\mathrm{2}{x}\right)+{b}^{{n}} \left(\mathrm{1}+{cos}\mathrm{2}{x}\right)}{{a}^{\mathrm{2}{n}} \left(\mathrm{1}−{cos}\mathrm{2}{x}\right)+{b}^{\mathrm{2}{n}} \left(\mathrm{1}+{cos}\mathrm{2}{x}\right)}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{a}^{{n}} +{b}^{{n}} −\left({a}^{{n}} −{b}^{{n}} \right){cos}\mathrm{2}{x}}{{a}^{\mathrm{2}{n}} +{b}^{\mathrm{2}{n}} −\left({a}^{\mathrm{2}{n}} −{b}^{\mathrm{2}{n}} \right){cos}\mathrm{2}{x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{A}−{Bcos}\mathrm{2}{x}}{{C}−{Dcos}\mathrm{2}{x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{A}}{{C}−{Dcos}\mathrm{2}{x}}{dx}+\frac{{B}}{{D}}\int_{\mathrm{0}} ^{\pi} \frac{{C}−{Dcos}\mathrm{2}{x}−{C}}{{C}−{Dcos}\mathrm{2}{x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{A}}{{C}−{Dcos}\mathrm{2}{x}}+\frac{{B}}{{D}}\int_{\mathrm{0}} ^{\pi} {dx}−\frac{{BC}}{{D}}\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{{C}−{Dcos}\mathrm{2}{x}} \\ $$$$\left({A}−\frac{{BC}}{{D}}\right)\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} {xdx}}{{C}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)−{D}\left(\mathrm{1}−{tan}^{\mathrm{2}} {x}\right)}+\frac{{B}}{{D}}\int_{\mathrm{0}} ^{\pi} {dx} \\ $$$$\left({A}−\frac{{BC}}{{D}}\right)\int_{\mathrm{0}} ^{\pi} \frac{{d}\left({tanx}\right)}{\left({C}−{D}\right)+\left({C}+{D}\right){tan}^{\mathrm{2}} {x}}+\frac{{B}}{{D}}\int_{\mathrm{0}} ^{\pi} {dx} \\ $$$$\frac{{AD}−{BC}}{{D}\left({C}+{D}\right)}\int_{\mathrm{0}} ^{\pi} \frac{{d}\left({tanx}\right)}{\frac{{C}−{D}}{{C}+{D}}+{tan}^{\mathrm{2}} {x}}+\frac{{B}}{{D}}\int_{\mathrm{0}} ^{\pi} {dx} \\ $$$$\frac{{AD}−{BC}}{{D}\left({C}+{D}\right)}×\frac{\mathrm{1}}{\sqrt{\frac{{C}−{D}}{{C}+{D}}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{tanx}}{\sqrt{\frac{{C}−{D}}{{C}+{D}}}}\right)\mid_{\mathrm{0}} ^{\pi} +\frac{{B}}{{D}}\pi \\ $$$$\frac{{B}}{{D}}\pi \\ $$$$=\frac{\left({a}^{{n}} −{b}^{{n}} \right)}{{a}^{\mathrm{2}{n}} −{b}^{\mathrm{2}{n}} }×\pi \\ $$$$=\frac{\pi}{{a}^{{n}} +{b}^{{n}} } \\ $$$${pls}\:{check}... \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 01/Apr/20

I love this. Great idea!!!

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

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

god bless you

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

Commented by TANMAY PANACEA. last updated on 01/Apr/20

thank you sir

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

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