zdebe5l8

2022-04-09

The question is, find the integral to the function:
$\frac{{\mathrm{sin}}^{3}\theta }{{\mathrm{sin}}^{3}\theta -{\mathrm{cos}}^{3}\theta }$

Gonarsu2dw8

$I=\int \frac{{\mathrm{sin}}^{3}\left(x\right)}{{\mathrm{sin}}^{3}\left(x\right)-{\mathrm{cos}}^{3}\left(x\right)}dx=\int \frac{{\mathrm{tan}}^{3}\left(x\right)}{{\mathrm{tan}}^{3}\left(x\right)-1}dx$
Let $t=\mathrm{tan}\left(x\right)$. Then $dt={\mathrm{sec}}^{2}\left(x\right)dx$. Since $\mathrm{tan}\left(x\right)=t$, we have ${\mathrm{sec}}^{2}\left(x\right)=1+{\mathrm{tan}}^{2}\left(x\right)=1+{t}^{2}$ and hence $dx=\frac{dt}{{\mathrm{sec}}^{2}\left(x\right)}=\frac{dt}{1+{t}^{2}}$
Now the integral becomes
$I=\int \frac{{t}^{3}}{\left({t}^{3}-1\right)\left(1+{t}^{2}\right)}dt$
Now resort to the good old partial fractions to get the integral.

folytonr4qx

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

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