from 0 to (π/2) ∫(sin^3θ cos^3θ dθ)
Math, not "college" - Writeacher, Monday, May 31, 2010 at 7:30am
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Integral calculus - MathMate, Monday, May 31, 2010 at 10:18am
When the sin/cos are in odd powers, use the substitution
∫sin³θcos³θd&theta (from 0 to π/2)
=∫(sin³θ-sin5θ)(d(sinθ)) (from 0 to sin(π/2))
=[sin4θ/4-sin6θ/6](from 0 to 1)