piarepm

2021-12-30

Evaluate the integral ${\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{3}x}{{\mathrm{sin}}^{3}x+{\mathrm{cos}}^{3}x}dx$

sonorous9n

As
${\int }_{a}^{b}f\left(x\right)dx={\int }_{a}^{b}f\left(a+b-x\right)dx$,
If
$I={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{n}x}{{\mathrm{sin}}^{n}x+{\mathrm{cos}}^{n}x}dx$
$={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{n}\left(\frac{\pi }{2}-x\right)}{{\mathrm{sin}}^{n}\left(\frac{\pi }{2}-x\right)+{\mathrm{cos}}^{n}\left(\frac{\pi }{2}-x\right)}dx$
$={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{cos}}^{n}x}{{\mathrm{cos}}^{n}x+{\mathrm{sin}}^{n}x}dx$
$⇒I+I={\int }_{0}^{\frac{\pi }{2}}dx$
assuming ${\mathrm{sin}}^{n}x+{\mathrm{cos}}^{n}x\ne 0$ which is true as $0\le x\le \frac{\pi }{2}$
Generalizition:
If
$Right\to owJ+J={\int }_{a}^{b}dx$
provided $g\left(x\right)+g\left(a+b-x\right)\ne 0$
If $a=0,b=\frac{\pi }{2}$ and $g\left(x\right)=h\left(\mathrm{sin}x\right)$,
$g\left(\frac{\pi }{2}+0-x\right)=h\mathrm{sin}\left(\frac{\pi }{2}+0-x\right)\right)=h\left(\mathrm{cos}x\right)$
So J becomes

Corgnatiui

Symmetry! This is the same as the integral with ${\mathrm{cos}}^{3}x$ on top.
If that is not obvious from the geometry, make the change of variable $u=\frac{\pi }{2}-x$
Add them, you get the integral of 1.. So our integral is $\frac{\pi }{4}$.

Vasquez

Hint: if
$I={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{3}x}{{\mathrm{sin}}^{3}x+{\mathrm{cos}}^{3}x}dx$
and
$J={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{cos}}^{3}x}{{\mathrm{sin}}^{3}x+{\mathrm{cos}}^{3}x}dx$
Then consider $I+J$, and the effect of the substitution $y=\frac{\pi }{2}-x$ on the integral I.

Do you have a similar question?