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Calculate the following integral by hands: $$\int_{0}^{\frac{\pi}{2}}\frac{\sin^{3}(t)}{\sin^{3}(t)+\cos^{3}(t)}dt$$

It seems apparently that integration by part does not work. I think integration by substitution could work but I can not figure out how to substitute.

I have calculated it by Wolfram and the result is $\frac{\pi}{4}$.

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  • $\begingroup$ I think Weierstrass substitution might help. $\endgroup$ – glebovg Oct 23 '12 at 21:44
  • $\begingroup$ If you needed the indefinite intregral, you could divide by $\cos^3t$ and substitute $t=\tan^{-1}x$ or divide by $\cos^5t$ and substitute $x=\tan t$. $\endgroup$ – Mike Oct 23 '12 at 22:12
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Write: $$\frac{\sin^3 t}{\sin^3 t+\cos^3 t} = 1-\frac{\cos^3 t}{\sin^3t+\cos^3t}$$

Apply the symmetry $\sin t = \cos(\frac\pi2 - t)$ to conclude that the integrals over the fractions are equal. The result from Wolfram follows.

While conceptually different, the use of symmetry is similar in effect to the substitution proposed by @Norbert.

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  • $\begingroup$ I'm pretty sure symmetry is just what Norbert had in mind, even though he didn't use that word. $\endgroup$ – Michael Hardy Oct 23 '12 at 20:57
  • $\begingroup$ @Lord_Farin I have followed your solution but what's next? I found that it is still difficult to find the indefinite integral and then solve the definite integral. $\endgroup$ – John Hass Oct 24 '12 at 16:53
  • $\begingroup$ When I wrote "integrals over the fractions" I meant the definite integrals. The indefinite integrals of these two fractions can't be equal obviously. If this comment doesn't make sense to you, it means I misunderstood your question. $\endgroup$ – Lord_Farin Oct 24 '12 at 17:03
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Hint: Make change of varianles $$t=\frac{\pi}{2}-x$$

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    $\begingroup$ You can actually show what $\int\limits_0^{\frac{\pi }{2}} {\dfrac{{{{\sin }^n}x}}{{{{\sin }^n}x + {{\cos }^n}x}}dx} $ is for every $n$ with the very same substitution. $\endgroup$ – Pedro Tamaroff♦ Oct 23 '12 at 17:44
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    $\begingroup$ Maybe you could also generalize this $\int\limits_0^{\frac{\pi }{2}} {\dfrac{{f({{\sin }}x)}}{{f({{\sin }}x) + f({{\cos }}x)}}dx}$ $\endgroup$ – clark Oct 23 '12 at 17:53

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