Cosine Conundrum: Cracking the Berkeley Bee Integral with

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Today, we're diving into a tricky little integral that popped up in the qualifying rounds of the Berkeley Integration Bee (HT @blackpenredpen for the problem). This is a classic example of how the hard part of calculus is almost never the calculus, but instead the algebra (and in this case a couple trig identities).

We start by dissecting the integrand using the difference of squares, a neat algebra trick. This transforms our expression into ((cos(x))^2 - (sin(x))^2)((cos(x))^2 + (sin(x))^2). The beauty lies in the simplification: the second part equals 1, thanks to the trusty Pythagorean identity. And the first part? It elegantly turns into cos(2x) with the double angle identity.

The crux of the solution lies in a simple u-substitution, steering us towards a surprisingly neat answer: 1/2. It's a reminder that sometimes, the most elegant solutions in mathematics come from stepping back and looking at the problem from a different angle.

#calculus #integrationbee #trigonometry

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