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

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4  
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
2  
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

4 Answers 4

up vote 6 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 $\cos^3x$

$$\int\frac{\tan^3xdx}{1+\tan^3x}$$

Now make the substitution

$$x=\tan^{-1}u,dx=\frac{du}{1+u^2}$$

$$\int\frac{u^3du}{(1+u^2)(1+u)(1-u+u^2)}$$

at which point it can be solved by partial fractions.

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1  
Very nice answer. Thanks. –  Legendre Sep 17 '12 at 18:00

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

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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.

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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 $I$.

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