∫√cos2Θ sin2ΘdΘ
let: u = cos2Θ
then: du/dΘ = -2sin2Θ
so: du = -2sin2ΘdΘ
so: du/ (-2sin2Θ) = dΘ
∫√cos2Θ sin2ΘdΘ = ∫√u sin2ΘdΘ =
∫√u sin2Θ(du/-2sin2Θ)
= -(1/2)∫√udu = (-1/2)*(2/3)u^(3/2)
= -1/3(cos2Θ)^(3/2)
â«âcos2Î sin2ÎdÎ
=âcos2Î sin2Î =â2 sin 4x
Try the substitution:
u = cos2Î
du/dx = -sin2Î
(-1/2)du = sin2ÎdÎ
Hope that helps!
is that the square root of everything?
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Verified answer
∫√cos2Θ sin2ΘdΘ
let: u = cos2Θ
then: du/dΘ = -2sin2Θ
so: du = -2sin2ΘdΘ
so: du/ (-2sin2Θ) = dΘ
∫√cos2Θ sin2ΘdΘ = ∫√u sin2ΘdΘ =
∫√u sin2Θ(du/-2sin2Θ)
= -(1/2)∫√udu = (-1/2)*(2/3)u^(3/2)
= -1/3(cos2Θ)^(3/2)
â«âcos2Î sin2ÎdÎ
=âcos2Î sin2Î =â2 sin 4x
Try the substitution:
u = cos2Î
du/dx = -sin2Î
(-1/2)du = sin2ÎdÎ
Hope that helps!
is that the square root of everything?