from 0 to (π/2) ∫(sin^3θ cos^3θ dθ)
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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
To evaluate the definite integral ∫(sin^3(θ) * cos^3(θ)) dθ over the interval [0, π/2], we can use trigonometric identities and integration techniques.
Let's start by using the product-to-sum trigonometric identity:
sin^3(θ) * cos^3(θ) = (sin(θ) * cos(θ))^3 = (1/2 * sin(2θ))^3.
Now, we have transformed the integral into:
∫(1/8 * sin^3(2θ)) dθ.
Next, we can use a u-substitution to simplify the integral further. Let u = 2θ, which implies du = 2dθ.
When θ = 0, u = 2(0) = 0.
When θ = π/2, u = 2(π/2) = π.
Therefore, the integral becomes:
∫(1/8 * sin^3(u)) * (1/2)du.
Simplifying, we get:
(1/16) * ∫sin^3(u) du.
To evaluate this integral, we can use a reduction formula. The reduction formula for sin^n(u) is:
∫sin^n(u) du = -1/n * sin^(n-1)(u) * cos(u) + (n-1)/n * ∫sin^(n-2)(u) du.
Applying the reduction formula to our integral:
∫sin^3(u) du = -1/3 * sin^2(u) * cos(u) + 2/3 * ∫sin(u) du.
We can evaluate the second term of the integral, ∫sin(u) du, which is equal to -cos(u):
= -1/3 * sin^2(u) * cos(u) + 2/3 * (-cos(u)).
Next, replace u with 2θ:
= -1/3 * sin^2(2θ) * cos(2θ) - 2/3 * cos(2θ).
Now, we need to evaluate this expression over the interval [0, π/2]:
= (-1/3 * sin^2(2(π/2)) * cos(2(π/2))) - 2/3 * cos(2(π/2)) - (-1/3 * sin^2(2(0)) * cos(2(0))) - 2/3 * cos(2(0)).
Simplifying further:
= (-1/3 * sin^2(π) * cos(π)) - 2/3 * cos(π) - (-1/3 * sin^2(0) * cos(0)) - 2/3 * cos(0).
Since sin(π) = 0 and cos(π) = -1, and sin(0) = 0 and cos(0) = 1, we can substitute these values:
= (-1/3 * (0)^2 * (-1)) - 2/3 * (-1) - (-1/3 * (0)^2 * (1)) - 2/3 * (1).
This simplifies to:
= 0 + 2/3 - 0 - 2/3.
Therefore, the value of the integral ∫(sin^3(θ) * cos^3(θ)) dθ over [0, π/2] is 0.
Note: The process of solving this definite integral involves a series of mathematical steps and techniques. It's important to have a solid understanding of trigonometric identities, integration rules, and substitution methods in order to accurately evaluate more complex integrals.