To evaluate the integral ∫cos^6(θ) dθ from 0 to π, we can use a trigonometric identity and then apply the power-reducing formula.
1. Start by using the identity cos^2(θ) = (1 + cos(2θ))/2 twice:
cos^6(θ) = (cos^2(θ))^3 = ((1 + cos(2θ))/2)^3.
2. Expand the expression:
cos^6(θ) = (1 + cos(2θ))^3/8.
3. Next, apply the power-reducing formula:
cos^6(θ) = (1 + 3cos(2θ) + 3cos^2(2θ) + cos^3(2θ))/8.
4. Simplify the expression inside the integral:
cos^6(θ) = (1 + 3cos(2θ) + 3(1 + cos(4θ))/2 + (1 + cos(2θ))^3/8)/8.
5. Distribute and combine like terms:
cos^6(θ) = (1 + 3cos(2θ) + 3/2 + 3cos(4θ)/2 + (1/8)(1 + 3cos(2θ) + 3cos^2(2θ) + cos^3(2θ)))/8.
6. Now, you can integrate each term separately.
The integral of 1 with respect to θ from 0 to π is π.
The integral of 3cos(2θ) with respect to θ from 0 to π is 0 (since cos(2θ) is an odd function over that interval).
The integral of 3/2 with respect to θ from 0 to π is (3/2)(π - 0) = (3/2)π.
The integral of 3cos(4θ)/2 with respect to θ from 0 to π is 0 (since cos(4θ) is an odd function over that interval).
Now, for the final term, we'll apply the power-reducing formula again:
The integral of (1 + 3cos(2θ) + 3cos^2(2θ) + cos^3(2θ))/8 with respect to θ from 0 to π is
[(θ + (3/2)sin(2θ) + (3/4)θ + (1/16)sin(4θ))/8] evaluated from 0 to π.
7. Evaluate the expression at the upper limit (π) and the lower limit (0):
[(π + (3/2)sin(2π) + (3/4)π + (1/16)sin(4π))/8] - [(0 + (3/2)sin(2(0)) + (3/4)(0) + (1/16)sin(4(0)))/8].
Since sin(2π) = 0 and sin(4π) = 0, these terms will simplify further:
[(π + (3/4)π)/8] - [(0)/8].
Simplifying, we get:
(4π/8) - (0/8) = π/2.
Therefore, the value of the integral ∫cos^6(θ) dθ from 0 to π is π/2.
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