Answer: Integral of sin³x/(sin³x+cos³x)

Key Idea / Intuition

The trick is to pair the integrand with its "complement" obtained by the substitution $x \mapsto \pi/2 - x$. This swap exchanges $\sin x \leftrightarrow \cos x$, so the integrand flips to its own "missing piece." Adding the two versions gives the constant function $1$, and therefore the integral over $[0,\pi/2]$ must be exactly half the length of the interval — no calculation needed.

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Formal Proof / Solution

Step 1: Apply the complementary substitution.

Let $x \mapsto \frac{\pi}{2} - x$. Since $dx \mapsto -dx$ and the limits swap back, we get

$I = \int_0^{\pi/2} \frac{\sin^3\!\left(\frac{\pi}{2}-x\right)}{\sin^3\!\left(\frac{\pi}{2}-x\right)+\cos^3\!\left(\frac{\pi}{2}-x\right)}\,dx = \int_0^{\pi/2} \frac{\cos^3 x}{\cos^3 x + \sin^3 x}\,dx.$

Call this $I'$. Note $I' = I$ is not needed; we just add the two expressions.

Step 2: Add $I$ and $I'$.

$I + I' = \int_0^{\pi/2} \left(\frac{\sin^3 x}{\sin^3 x + \cos^3 x} + \frac{\cos^3 x}{\cos^3 x + \sin^3 x}\right)dx = \int_0^{\pi/2} 1\, dx = \frac{\pi}{2}.$

Step 3: Conclude.

Since both expressions equal $I$ (the substitution produced the same integral), we have $2I = \dfrac{\pi}{2}$, so

$\boxed{I = \frac{\pi}{4}.}$

Why this is elegant: The integrand $f(x) = \dfrac{\sin^3 x}{\sin^3 x + \cos^3 x}$ satisfies $f(x) + f\!\left(\tfrac{\pi}{2}-x\right) = 1$ for every $x$. Any integrand with this "complementary symmetry" on $[0,\pi/2]$ integrates to exactly $\pi/4$, regardless of the exponent. Indeed the same trick works for $\dfrac{\sin^n x}{\sin^n x + \cos^n x}$ for any $n > 0$.