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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?

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  • $\begingroup$ Does the integral happen to be from $0$ to $\frac{\pi}{2}$ $\endgroup$ – Teh Rod Feb 11 '17 at 16:02
  • $\begingroup$ @Teh Rod, No, it is indefinite. $\endgroup$ – user300045 Feb 11 '17 at 16:42
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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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    $\begingroup$ Was thinking of this method. Good job $\endgroup$ – lab bhattacharjee Feb 11 '17 at 14:03
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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 : $$I = \frac{x}{2} + \frac{\ln(\sin x+\cos x)-2\ln(2-\sin2x)}{6} + C$$

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$\displaystyle\frac{\cos^3x}{\sin^3x+\cos^3x}=\frac{1}{\tan^3x+1}=\frac{\tan^2x+1}{(\tan^2x+1)(\tan^3x+1)}=\frac{1}{\cos^2x(\tan^2x+1)(\tan^3x+1)}$

Now replace $\displaystyle t=\tan x,dt=\frac{dx}{\cos^2x}$ and do partial fractions.

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