Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
And you tried to split into partial fractions afterwards, didn't you? –  J. M. Nov 13 '11 at 2:37
add comment

3 Answers

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.

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

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).

share|improve this answer
    
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
add comment

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: $I+J=(1/3)\ln\left(\dfrac{|u|}{(3-u^2)^2}\right).$

share|improve this answer
    
+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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.