Use the Table of Integrals to evaluate the integral.
0 to π ∫cos^6(θ) dθ
I don't see anything like that on the table of integral. Need some help.
Yeah, you probably won't find 6th powers of trig functions in the table. Many go up to 4 or 5. What you have to rely on here is the double angle formulas to lower the power.
cos2x = 2cos^2x - 1
so, cos^2x = (1+cos2x)/2
so, cos^6x = (cos^2x)^3 = ((1+cos2x)/2)^3
= 1/8 (1+3cos(2x)+3cos^2(2x) + cos^3(2x))
You can probably find powers 1,2,3 of cos(x) in your table.
when you're all done, you can check your work with any of many handy integral calculors, such as wolframalpha.com
Just type in your integral
integral[0..pi] cos^6(θ) dθ
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.