Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is it possible to evaluate the following integral:$$\int \frac{\sin^3x}{(\sin^3x + \cos^3x)} \, dx$$

share|cite|improve this question
You can factor the denominator. Your answer contains log(sin(x)+cos(x)) maybe that will help you. – mick Sep 17 '12 at 16:10
@mick : I up-voted your comment, and then realized I still don't get it. The denominator becomes (a+b)(1-ab) using trig identity. But I am still kinda stuck. :( – Legendre Sep 17 '12 at 16:38
here's a detailed process ... it bet this can be much simpler. – Santosh Linkha Sep 17 '12 at 16:51
@experimentX : Nice, I didn't know wolfram alpha can do that. Well, at least we got the solution to try and reverse engineer from. – Legendre Sep 17 '12 at 16:54
up vote 11 down vote accepted

Well, I'm still not seeing any nice ways of doing it. I do see at least one way of proceeding though. First, divide top and bottom by $\cos^3x$


Now make the substitution



at which point it can be solved by partial fractions.

share|cite|improve this answer
Very nice answer. Thanks. – Legendre Sep 17 '12 at 18:00

Put $$I = \displaystyle\int{\dfrac{\sin^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x , \quad J = \displaystyle\int{\dfrac{\cos^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x.$$ We have $$I + J = \displaystyle\int{\mathrm{d}x} = x+ C.$$ and \begin{equation*} I - J = \displaystyle\int{\dfrac{\sin^3 x - \cos^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x = \displaystyle\int{\dfrac{(\sin x - \cos x)(1 + \sin x \cdot \cos x)}{(\sin x + \cos x)(1 - \sin x \cdot \cos x)}\mathrm{d}x} \end{equation*} Put $t = \sin x + \cos x$, then $\mathrm{d}t = -(\sin x - \cos x) \mathrm{d}x$ and $ \sin x \cdot \cos x = \dfrac{ t^2-1}{2}.$ We get \begin{equation*} I - J = \displaystyle\int{\dfrac{t^2 + 1}{t(t^2-3)}\mathrm{d}t} = \dfrac{2}{3}\ln{(t^2-3)}-\dfrac{1}{3}\ln t + C'. \end{equation*} and then \begin{equation*} I - J = \dfrac{2}{3}\ln{((\sin x + \cos x)^2-3)}-\dfrac{1}{3}\ln (\sin x + \cos x) + C'. \end{equation*} From $I + J $ and $I - J$, we can calculate $I$.

share|cite|improve this answer
this method looks smart – Ekaveera Kumar Sharma Oct 10 '15 at 2:36

If all else fails, the Weierstrass substitution will do it.

share|cite|improve this answer
I suppose that does the job. But is there a "simpler" method? E.g. factoring and simplifying? – Legendre Sep 17 '12 at 16:51

I think that if this exercise can be solved in a simple fashion (without heavy computation), then this should be the way to approach it (otherwise, it is just a mindless computation which just requires to apply some algorithm like the, so-called, Weierstrass substitution and teaches you nothing).

So, since $$ \sin^{3}x+\cos^{3}x=\left( \sin x+\cos x\right) \left( \sin^{2}x-\sin x\cos x+\cos^{2}x\right) =\left( \sin x+\cos x\right) \left( 1-\sin x\cos x\right) , $$ then we try to express $\dfrac{\sin^{3}x}{\sin^{3}x+\cos^{3}x}$ as follows (if possible), in order to be able to (easily) compute ${\displaystyle\int}\dfrac{\sin^{3}x}{\sin^{3}x+\cos^{3}x}\;\mathrm{d}x$: \begin{align*} \frac{\sin^{3}x}{\sin^{3}x+\cos^{3}x} & =A+B\cdot\frac{\left( \sin x+\cos x\right) ^{\prime}}{\sin x+\cos x}+C\cdot\frac{\left( 1-\sin x\cos x\right) ^{\prime}}{1-\sin x\cos x}=\\ & =A+B\cdot\frac{\cos x-\sin x}{\sin x+\cos x}+C\cdot\frac{\sin^{2}x-\cos ^{2}x}{1-\sin x\cos x} \end{align*} From here we obtain that $$ \sin^{3}x=A\left( \sin^{3}x+\cos^{3}x\right) +B\left( \cos x-\sin x\right) \left( 1-\sin x\cos x\right) +C\left( \sin x+\cos x\right) \left( \sin^{2}x-\cos^{2}x\right) $$ hence \begin{align*} 0 & =(A+C-1)\sin^{3}x+(A-C)\cos^{3}x+B\left( \cos x-\sin x\right) -(B+C)\sin x\cos^{2}x+(B+C)\sin^{2}x\cos x\\ & =(A+C-1)\sin^{3}x+(A-C)\cos^{3}x+B\left( \cos x-\sin x\right) -(B+C)\sin x(1-\sin^{2}x)+(B+C)(1-\cos^{2}x)\cos x\\ & =(A+B+2C-1)\sin^{3}x+(A-B-2C)\cos^{3}x+(2B+C)\left( \cos x-\sin x\right) \end{align*} so $$ \left\{ \begin{array} [c]{r} A+B+2C=1\\ A-B-2C=0\\ 2B+C=0 \end{array} \right. $$ which has the (unique) solution $$ \left\{ \begin{array} [c]{l} A=\frac{1}{2}\\ B=-\frac{1}{6}\\ C=\frac{1}{3} \end{array} \right. $$ hence \begin{align*} {\displaystyle\int}\dfrac{\sin^{3}x}{\sin^{3}x+\cos^{3}x}\;\mathrm{d}x &=Ax+B\log\left\vert \sin x+\cos x\right\vert +C\log\left\vert 1-\sin x\cos x\right\vert +\text{some constant}\\ &=\frac{x}{2}-\frac{\log\left\vert \sin x+\cos x\right\vert }{6}+\frac {\log\left\vert 1-\sin x\cos x\right\vert }{3}+\text{some constant} \end{align*} Let's hope I didn't make any mistake in my calculations.

share|cite|improve this answer

Your Answer


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.