# $\int \frac{\sin^3x}{\sin^3x + \cos^3x)}$?

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

#### Solutions Collecting From Web of "$\int \frac{\sin^3x}{\sin^3x + \cos^3x)}$?"

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.

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 $I$.

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

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.