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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
    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.2012-03-15

2 Answers 2

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

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

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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 $n$.