from 0 to (π/2) ∫(sin^3θ cos^3θ dθ)
2 answers
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When the sin/cos are in odd powers, use the substitution
cosθdθ=d(sinθ), or
sinθdθ=-d(cosθ)
∫sin³θcos³θd&theta (from 0 to π/2)
=∫sin³θ(1-sin²θ) (cosθdθ)
=∫(sin³θ-sin5θ)(d(sinθ)) (from 0 to sin(π/2))
=[sin4θ/4-sin6θ/6](from 0 to 1)
=1/12
cosθdθ=d(sinθ), or
sinθdθ=-d(cosθ)
∫sin³θcos³θd&theta (from 0 to π/2)
=∫sin³θ(1-sin²θ) (cosθdθ)
=∫(sin³θ-sin5θ)(d(sinθ)) (from 0 to sin(π/2))
=[sin4θ/4-sin6θ/6](from 0 to 1)
=1/12