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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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    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 $x=\tan t$.2012-10-23

2 Answers 2

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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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    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.2012-10-24
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Hint: Make change of varianles $t=\frac{\pi}{2}-x$

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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}$2012-10-23