# How can I calculate $\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}dt$?

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

• 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 $$\int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}dt$$ for $n>3$?
• [EDIT:] For the pedagogical reason, can anyone come up with a references for this exercise?
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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. –  Mike Mar 15 '12 at 14:05

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

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

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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 $n$.

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+1, and I've done some TeX corrections. Notice that when you write \cos x, getting $\cos x$, the backslash not only prevents italicization but also results in proper spacing, whereas cos x without the backslash, even with a space between "cos" and "x", looks like this: $cos x$. In some contexts, e.g. with \sup, \max, etc., when in a "displayed" as opposed to "inline" setting, the backslash also affects positioning of subscripts or superscripts. –  Michael Hardy Nov 16 '11 at 2:20
@MichaelHardy. Thanks Micheal. –  smanoos Nov 16 '11 at 2:47
@smanoos, thanks for your answer. I edited it in the way that I think may be more readable. –  Jack Nov 16 '11 at 3:37
@Jack. Thank you. –  smanoos Nov 16 '11 at 7:52