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Question Number 175042 by peter frank last updated on 17/Aug/22

Answered by Frix last updated on 18/Aug/22

$$\int \frac{a^2{\sin }^{2}x+b^2{\cos }^{2}x}{a^4{\sin }^{4}x+b^4{\cos }^{4}x}dx=$$$$=\int \frac{(1+{\tan }^{2}x)(b^2+a^2{\tan }^{2}x)}{b^4+a^4{\tan }^{4}x}dx=$$$$=\int \frac{b^2+a^2{\tan }^{2}x}{(b^4+a^4{\tan }^{4}x)}d\tan x=$$$$=\int \frac{a^2{t}^2+b^2}{a^4{t}^4+b^4}dt=$$$$=\frac{1}{2}\int \frac{dt}{a^2{t}^2-ab\sqrt{2}t+b^2}+\frac{1}{2}\int \frac{dt}{a^2{t}^2+ab\sqrt{2}t+b^2}=$$$$=\frac{\sqrt{2}}{2ab}({\tan }^{-1}\frac{a\sqrt{2}t-b}{b}+{\tan }^{-1}\frac{a\sqrt{2}t+b}{b})=$$$$=-\frac{\sqrt{2}}{2ab}{\tan }^{-1}\frac{ab\sqrt{2}t}{a^2{t}^2-b^2}=$$$$=-\frac{\sqrt{2}}{2ab}{\tan }^{-1}\frac{ab\sqrt{2}\sin x\cos x}{a^2{\sin }^{2}x-b^2{\cos }^{2}x}+C$$

Commented by peter frank last updated on 18/Aug/22

$$\mathrm{thank}\mathrm{you}$$

Commented by Tawa11 last updated on 20/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

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