<s>[INST] Solve an integral $\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$ Solve an integral $$\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$$
I tried to divide the numerator and denominator by $\cos^4 x$ to get $\sec x$ function but the term ${\sin^3 x}/{\cos^4 x}$ gives $\tan^2 x\sec^2 x\sin x$. How to get rid of $\sin x$ term?
 [/INST] I wasn't really able to come up with a better (elegant) method other than the following:
$$\int \frac{\cos^3 x}{\sin^3 x + \cos^3 x} \mathrm{d}x = \int \frac{1}{1 + \tan^3 x} \mathrm{d}x$$
Now, using the substitution, $t = \tan x \implies \frac{\mathrm{d}t}{1+t^2} = \mathrm{d}x$, we get
$$= \int \frac{1}{(1 + t^2)(1+t^3)} \mathrm{d}t$$
Decomposing it into partial fraction (copying from W|A): 
$$= \int \frac{1}{6(t+1)} + \frac{t+1}{2(t^2+1)} - \frac{2t-1}{3(t^2-t+1)} \mathrm{d}t \\ = \frac 16\ln t + \frac 14\ln (t^2+1) + \frac 12\arctan t - \frac 13 \ln (t^2-t+1) + C$$
Substituting back $t = \tan x$
$$\frac 16 \ln \tan x + \frac 12 \ln \sec x -\frac 13 \ln (\sec^2x - \tan x) + \frac x2 + C$$
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