2006 SMT/Calculus Problems/Problem 6

Solution

Let $u=\frac{\pi}{2}-x$ and $du=-dx$ , the integral than turns into:

$I = \int_{0}^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx \Rightarrow -\int_{\pi/2}^{0}\frac{\sin^3(\pi/2-u)}{\sin^3(\pi/2-u)+\cos^3(\pi/2-u)}du$

Which of course equals: $I =-\int_{\pi/2}^{0}\frac{\sin^3(\pi/2-u)}{\sin^3(\pi/2-u)+\cos^3(\pi/2-u)}du = \int_{0}^{\pi/2}\frac{\cos^3(u)}{\cos^3(u)+\sin^3(u)}du$

We can then add the original integral, with x as the variable, to the integral with u as a variable because the variables don't matter, just the value that the integrals takes:

$\begin{align*} I+I&=\int_{0}^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx+\int_{0}^{\pi/2}\frac{\cos^3x}{\cos^3x+\sin^3x}dx\\ 2I&=\int_{0}^{\pi/2}\frac{\sin^3x+\cos^3x}{\sin^3x+\cos^3x}dx\\ 2I&=\int_{0}^{\pi/2}1dx\\ 2I&=x\Big|_0^{\frac{\pi}{2}}\\ 2I&=\frac{\pi}{2}\\ I&=\frac{\pi}{4}\\ \end{align*}$

Therefore:

$I=\int_{0}^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx = \boxed{\frac{\pi}{4}}$

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2006 SMT/Calculus Problems/Problem 6

Solution