October
23, 2006
Evaluate
$\displaystyle\int_0^{\pi/2}
\frac{\sin^{2006}x}{\sin^{2006}x+\cos^{2006}x}\,dx$
Hint: If $f$ is a
continuous function, then $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$.
Solution. Let $I$ be the
given integral. Using the hint with $f(x)$ being the integrand, and
$a=\pi/2$, we have
$\displaystyle I=\int_0^{\pi/2}
\frac{\sin^{2006}(\pi/2 - x)}{\sin^{2006}(\pi/2-x) +
\sin^{2006}(\pi/2-x)}$.
We know, however that $\sin (\pi/2-x)=\cos x$ and $\cos (\pi/2-x)=\sin
x$. Thus,
$I=\displaystyle\int_0^{\pi/2}
\frac{\cos^{2006}x}{\cos^{2006}x+\sin^{2006}x}\,dx$
and
$2I=\displaystyle\int_0^{\pi/2}
\frac{\sin^{2006}x}{\sin^{2006}x+\cos^{2006}x}\,dx + \int_0^{\pi/2}
\frac{\cos^{2006}x}{\cos^{2006}x+\sin^{2006}x}\,dx = \int_0^{\pi/2}
\frac{\sin^{2006}x+\cos^{2006}x}{\sin^{2006}x+\cos^{2006}x}\,dx =
\int_0^{\pi/2}dx = \frac{\pi}2$.
Thus $I=\pi/4$.