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

from 0 to (π/2) ∫(sin^3⁡θ cos^3⁡θ dθ)

  • Math, not "college" - , 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 - , 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-sin5θ)(d(sinθ)) (from 0 to sin(π/2))
    =[sin4θ/4-sin6θ/6](from 0 to 1)
    =1/12

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