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

$${\int }_0^{\pi }\frac{a^n{sin}^2\left(x\right)+b^n{cos}^2\left(x\right)}{a^{2n}{sin}^2\left(x\right)+b^{2n}{cos}^2\left(x\right)}dx;a>b$$

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

$${\int }_0^{\pi }\frac{a^n(1-cos2x)+b^n(1+cos2x)}{a^{2n}(1-cos2x)+b^{2n}(1+cos2x)}dx$$$${\int }_0^{\pi }\frac{a^n+b^n-(a^n-b^n)cos2x}{a^{2n}+b^{2n}-(a^{2n}-b^{2n})cos2x}dx$$$${\int }_0^{\pi }\frac{A-Bcos2x}{C-Dcos2x}dx$$$${\int }_0^{\pi }\frac{A}{C-Dcos2x}dx+\frac{B}{D}{\int }_0^{\pi }\frac{C-Dcos2x-C}{C-Dcos2x}dx$$$${\int }_0^{\pi }\frac{A}{C-Dcos2x}+\frac{B}{D}{\int }_0^{\pi }dx-\frac{BC}{D}{\int }_0^{\pi }\frac{dx}{C-Dcos2x}$$$$(A-\frac{BC}{D}){\int }_0^{\pi }\frac{{sec}^{2}xdx}{C(1+{tan}^{2}x)-D(1-{tan}^{2}x)}+\frac{B}{D}{\int }_0^{\pi }dx$$$$(A-\frac{BC}{D}){\int }_0^{\pi }\frac{d\left(tanx\right)}{(C-D)+(C+D){tan}^{2}x}+\frac{B}{D}{\int }_0^{\pi }dx$$$$\frac{AD-BC}{D(C+D)}{\int }_0^{\pi }\frac{d\left(tanx\right)}{\frac{C-D}{C+D}+{tan}^{2}x}+\frac{B}{D}{\int }_0^{\pi }dx$$$$\frac{AD-BC}{D(C+D)}\times \frac{1}{\sqrt{\frac{C-D}{C+D}}}\mid {tan}^{-1}\left(\frac{tanx}{\sqrt{\frac{C-D}{C+D}}}\right){\mid }_0^{\pi }+\frac{B}{D}\pi$$$$\frac{B}{D}\pi$$$$=\frac{(a^n-b^n)}{a^{2n}-b^{2n}}\times \pi$$$$=\frac{\pi }{a^n+b^n}$$$$plscheck...$$$$$$

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

$$Ilovethis.Greatidea!!!$$$$$$

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

$$godblessyou$$

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

$$thankyousir$$

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