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Question Number 138329 by JulioCesar last updated on 12/Apr/21

Answered by Ñï= last updated on 12/Apr/21

$$\int \frac{dx}{\sin {}^{4}x+\cos {}^{4}x}=\int \frac{dx}{1-2\sin {}^{2}x\cos {}^{2}x}=\int \frac{dx}{1-\frac{1}{2}{\sin }^{2}2x}$$$$=\int \frac{{csc}^{2}2xd}{{csc}^{2}2x-\frac{1}{2}}=-\int \frac{d\left(cot2x\right)}{2{cot}^{2}2x+1}=-{\tan }^{-1}\left(\sqrt{2}\cot 2x\right)+C$$

Commented by JulioCesar last updated on 12/Apr/21

$$thanksir!$$

Answered by Ñï= last updated on 12/Apr/21

$$\int \frac{dx}{\sin {}^{4}x+\cos {}^{4}x}=\int \frac{\tan {}^{2}x+1}{\tan {}^{4}x+1}d\left(\tan x\right)$$$$=\int \frac{u^2+1}{u^4+1}du=\int \frac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}}du=\int \frac{d(u-\frac{1}{u})}{{(u-\frac{1}{u})}^2+2}$$$$=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{u-\frac{1}{u}}{\sqrt{2}}+C=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{\tan x-\cot x}{\sqrt{2}}+C$$

Answered by MJS_new last updated on 12/Apr/21

$$\int \frac{dx}{{\sin }^{4}x+{\cos }^{4}x}=$$$$[t=\tan 2x\to dx=\frac{1}{2}{\cos }^{2}2xdt]$$$$=\int \frac{dt}{t^2+2}=\frac{\sqrt{2}}{2}\arctan \frac{\sqrt{2}t}{2}=\frac{\sqrt{2}}{2}\arctan \frac{\sqrt{2}\tan 2x}{2}+C$$

Commented by JulioCesar last updated on 12/Apr/21

$$Thanksir!$$

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