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?
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 $t = \tan x$
$$\frac 16 \ln \tan x + \frac 12 \ln \sec x -\frac 13 \ln (\sec^2x - \tan x) + \frac x2 + C$$
Let $$I = \int \frac{\cos^3x}{\sin^3x+\cos^3x} dx$$ $$I_1 = \int\frac{\sin^3x+\cos^3x}{\sin^3x+cos^3x}dx = x + C$$ and $$I_2 = \int\frac{\cos^3x-\sin^3x}{\sin^3x+\cos^3x}dx$$Then $$I = \frac{I_1 + I_2}{2}$$ $$I_2 = \int\frac{(\cos x-\sin x)(1+\frac{\sin2x}{2})}{(\sin x+\cos x)(1-\frac{\sin 2x}{2})}dx$$ Now substitute $$t = \sin x+\cos x$$ and you get $$dt = (\cos x- \sin x)dx$$ and $$\frac{\sin2x}{2} = \frac{t^2-1}{2}$$ Then $$I_2 = \int\frac{t^2+1}{t(3-t^2)}dt = \frac{\ln t - 2\ln(3-t^2)}{3} + C$$ Finally, when putting this back together : $$I = \frac{x}{2} + \frac{\ln(\sin x+\cos x)-2\ln(2-\sin2x)}{6} + C$$
$\displaystyle\frac{\cos^3x}{\sin^3x+\cos^3x}=\frac{1}{\tan^3x+1}=\frac{\tan^2x+1}{(\tan^2x+1)(\tan^3x+1)}=\frac{1}{\cos^2x(\tan^2x+1)(\tan^3x+1)}$
Now replace $\displaystyle t=\tan x,dt=\frac{dx}{\cos^2x}$ and do partial fractions.