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The question is, find the integral to the function:

$\sin^3\theta / (\sin^3\theta - \cos^3\theta)$

The only thing I could think of was to factor the denominator. But then I couldn't make any further progress.

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And you tried to split into partial fractions afterwards, didn't you? – J. M. Nov 13 '11 at 2:37
up vote 11 down vote accepted

$\displaystyle I = \int \frac{\sin^{3}(x)}{\sin^{3}(x) - \cos^{3}(x)}dx = \int \frac{\tan^{3}(x)}{\tan^{3}(x) - 1}dx $

Let $t = \tan(x)$. Then $dt = \sec^2(x) dx$. Since $\tan(x) = t$, we have $\sec^2(x) = 1 + \tan^2(x) = 1 + t^2$ and hence $dx = \frac{dt}{\sec^2(x)} = \frac{dt}{1+t^2}$.

Now the integral becomes $$ \displaystyle I = \int \frac{t^3}{(t^3-1)(1+t^2)}dt$$

Now resort to the good old partial fractions to get the integral.

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Can you please elaborate on $dt = \sec^2(x) dx \implies dx = \frac{dt}{1+t^2}$? – Mark Nov 13 '11 at 3:26
$\sec^2(x) = \tan^2(x) + 1 = t^2 + 1$, so $dt = (t^2 + 1) dx$ and $\frac{dt}{t^2+1} = dx$. – Patrick Da Silva Nov 13 '11 at 3:32

We take advantage of the symmetry, indeed expand on it. Let $$I=\int \frac{\sin^3\theta\,d\theta}{\sin^3\theta-\cos^3\theta} \qquad\text{and}\qquad J=\int \frac{\cos^3\theta\,d\theta}{\sin^3\theta-\cos^3\theta}.$$ Note that $$\frac{\sin^3\theta}{\sin^3\theta-\cos^3\theta}=1+ \frac{\cos^3\theta}{\sin^3\theta-\cos^3\theta},$$ and therefore $$I-J=\theta.$$ If we can find $I+J$ we will be finished. So we want to find $$\int\frac{\sin^3\theta+\cos^3\theta}{\sin^3\theta-\cos^3\theta}\,d\theta= \int\frac{(\sin\theta+\cos\theta)(\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta)}{(\sin\theta-\cos\theta)(\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta) }\,d\theta.$$ Let $u=\sin\theta-\cos\theta$. Then $du=(\cos\theta+\sin\theta)\,d\theta$. Also, $u^2=1-2\sin\theta\cos\theta$. From this we find that $\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta=\frac{1+u^2}{2}$ and $\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta=\frac{3-u^2}{2}$. Thus $$I+J=\int\frac{1+u^2}{u(3-u^2)}\,du.$$ We do a partial partial fraction decomposition:
$$\frac{1+u^2}{u(3-u^2)}=\frac{1}{3}\left(\frac{1}{u}+\frac{4u}{3-u^2}\right).$$ Integrate: $I+J=(1/3)\ln\left(\dfrac{|u|}{(3-u^2)^2}\right).$

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+1, symmetrization. – zyx Nov 13 '11 at 18:07
And then add on $/theta$ and divide the whole thing by 2, correct? – Mark Nov 13 '11 at 21:58
Yes, we know $I-J$ and $I+J$, so add them, divide by $2$. And remember the $+C$ at the end! – André Nicolas Nov 13 '11 at 22:16

If all else fails, the Weierstrass substitution should do things like this, but only if you can tolerate messy algebraic equations requiring numerical solutions. Let's try it: $$ \begin{align} & {} \qquad \int \frac{\Big(2t/(1+t^2)\Big)^3}{\Big(2t/(1+t^2)\Big)^3 - \Big((1-t^2)/(1+t^2)\Big)^3} \; \frac{2\;dt}{1+t^2} \\ \\ \\ & = \int \frac{(2t)^3}{(2t)^3 - (1-t^2)^3} \; \frac{2\;dt}{1+t^2} \\ \\ \\ & = \int\frac{8t^3}{t^6 - 3t^4 + 8t^3 + 3t^2 -1} \; \frac{2\;dt}{1+t^2}. \end{align} $$ At this point I think you'd have to result to numerical methods to factor the thing, but it looks as if the sixth-degree polynomial would be the product of two distinct first-degree factors and two irreducible quadratic factors. When you then find the partial fraction decomposition, there'd be the question of where you get $\text{constant}/\text{irreducible quadratic}$ (so you'd get an arctangent) and where you get $t/\text{irreducible quadratic}$ (so you'd get a logarithm).

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The degree 6 polynomial factorizes rationally as $(t^2+2 t-1) (t^4-2 t^3+2 t^2+2 t+1)$. To split it completely you need to adjoin some square roots. – zyx Nov 13 '11 at 18:51

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