Posted by **Laplace transform** on Monday, May 31, 2010 at 6:59am.

from 0 to (π/2) ∫(sin^3θ cos^3θ dθ)

- Math, not "college" -
**Writeacher**, Monday, May 31, 2010 at 7:30am
Please type your __subject__ in the **School Subject** box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.

- Integral calculus -
**MathMate**, Monday, May 31, 2010 at 10:18am
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-sin^{5}θ)(d(sinθ)) (from 0 to sin(π/2))

=[sin^{4}θ/4-sin^{6}θ/6](from 0 to 1)

=1/12

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