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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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I think Weierstrass substitution might help. –  glebovg Oct 23 '12 at 21:44
    
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 $x=\tan t$. –  Mike Oct 23 '12 at 22:12

2 Answers 2

up vote 5 down vote accepted

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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I'm pretty sure symmetry is just what Norbert had in mind, even though he didn't use that word. –  Michael Hardy Oct 23 '12 at 20:57
    
@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. –  John Hass Oct 24 '12 at 16:53
    
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. –  Lord_Farin Oct 24 '12 at 17:03

Hint: Make change of varianles $$t=\frac{\pi}{2}-x$$

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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. –  Pedro Tamaroff Oct 23 '12 at 17:44
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Maybe you could also generalize this $\int\limits_0^{\frac{\pi }{2}} {\dfrac{{f({{\sin }}x)}}{{f({{\sin }}x) + f({{\cos }}x)}}dx}$ –  clark Oct 23 '12 at 17:53

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