Question

from 0 to (π/2) ∫(sin^3⁡θ cos^3⁡θ dθ)

Answer 1

Please type your <u>subject</u> in the <b>School Subject</b> box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.

Answer 2

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