Calculating with Mathematica, one can have $$\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}\,\mathrm dt=\frac{\pi}{4}.$$

  • How can I get this formula by hand? Is there any simpler idea than using $u = \sin t$?
  • Is there a simple way to calculate $$ \int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}\,\mathrm dt $$ for $n>3$?
  • Could anyone come up with a reference for this exercise?
  • 1
    $\begingroup$ If I had to solve the indefinite case, I'd probably divide top and bottom by $\cos^3t$, make the substitution $t=\tan^{-1}x$, then evaluate with partial fractions. $\endgroup$ – Mike Mar 15 '12 at 14:05

The substitution $y=\frac{\pi}{2}-t$ solves it... If you do this substitution, you get:

$$\int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}dt= \int_0^{\pi/2}\frac{\cos^n y}{\cos^n y+\sin^n y}dy \,.$$


Use the Calculus identity that $$f(x)=f(a-x),$$ and let $$I=\int_0^\frac{\pi}{2} \frac{\sin^3t}{\sin^3t+\cos^3t}dt.$$ Then, $$f(t)=f(\frac{\pi}{2}-t)=\frac{\sin^3(\frac{\pi}{2}-t)}{\sin^3(\frac{\pi}{2}-t)+\cos^3(\frac{\pi}{2}-t)}=\frac{\cos^3t}{\cos^3t+\sin^3t}$$Thus, $$I=\int_0^\frac{\pi}{2} \frac{\cos^3t}{\cos^3t+\sin^3t}dt.$$ So we have $$2I=\int_0^\frac{\pi}{2} \frac{\sin^3t}{\sin^3t+\cos^3t}dt+\int_0^\frac{\pi}{2} \frac{\cos^3t}{\cos^3t+\sin^3t}dt=\int_0^\frac{\pi}{2}dt=\frac{\pi}{2}.$$ So $$I=\frac{\pi}{4}.$$ Note that this is true for any natural number $n$.


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