<s>[INST] Evaluate $\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$ $$\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$$
$$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$
$$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$
$$\int \frac{(1- \cos2x)^2}{2.(1+\cos^2 2x)}{dx}$$
$$\frac{1}{2} \int \left[1-\frac{2 \cos2x}{1+\cos^22x}\right] dx$$
What should I do next ?
Please also tell me alternative way to do this .
 [/INST] $$I=\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}=\int \frac{1-2\cos^2(x)+\cos^4(x)}{1-2\cos^2(x)+2\cos^4(x)}dx$$
$$=\int\frac{1}{2}-\frac 12\frac{-1+2\cos^2(x)}{1-2\cos^2(x)+2\cos^4(x)}dx$$
$$=\int\frac{1}{2}-\frac 12\frac{\cos(2x)}{1-2\cos^2(x)\sin^2(x)}dx$$
$\color{blue}{\text{This last step above is a few manipulations away from being the last step in your question.}}$
Let $u=\cos(x)\sin(x)$ then $du=\cos(2x)dx$
$$\boxed{I=\frac x2-\frac 12\int \frac 1{1-2u^2}du=\frac x2 -\frac12\frac{\tanh^{-1}(\sqrt2 u)}{\sqrt 2} +k_1}$$
Replace $u$ and you're good to go...
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