Express sin^6theta in multiples of costheta and hence evaluate the integral of sin^6theta from 0 to pi/2.
Thanks in advance! :)
sin^6θ = (sin^2θ)^3
= (1 - cos^2θ)^3
= 1 - 3cos^2θ + 3cos^4θ + cos^6θ
Now you can use your half-angle formula
cos^2θ = (1 + cos 2θ)/2
to get no exponents and multiples of θ.
You will end up with
1/192 (60θ - 45sin2θ + 9sin4θ - sin6θ)
from 0 to pi/2 yields 30pi/192
To express sin^6(theta) in terms of cos(theta), we can use the identity sin^2(theta) + cos^2(theta) = 1. Rearranging this identity, we have sin^2(theta) = 1 - cos^2(theta).
Now, let's substitute sin^2(theta) using this identity in sin^6(theta):
sin^6(theta) = (sin^2(theta))^3 = (1 - cos^2(theta))^3.
Expanding this using the binomial theorem, we have:
sin^6(theta) = (1 - cos^2(theta))^3 = 1 - 3cos^2(theta) + 3cos^4(theta) - cos^6(theta).
Now, let's evaluate the integral of sin^6(theta) from 0 to pi/2:
∫[0 to π/2] sin^6(theta) d(theta)
= ∫[0 to π/2] (1 - 3cos^2(theta) + 3cos^4(theta) - cos^6(theta)) d(theta)
To integrate this expression, we can use the power-reducing formula for cos^2(theta) and cos^4(theta):
∫[0 to π/2] (1 - 3cos^2(theta) + 3cos^4(theta) - cos^6(theta)) d(theta)
= ∫[0 to π/2] (1 - 3cos^2(theta) + 3cos^4(theta) - cos^6(theta)) d(theta)
= ∫[0 to π/2] (1 - 3cos^2(theta) + 3cos^4(theta) - cos^6(theta)) d(theta)
Using the power-reducing formulas for cos^2(theta) and cos^4(theta), we can rewrite the integral as:
∫[0 to π/2] (1 - 3cos^2(theta) + 3cos^4(theta) - cos^6(theta)) d(theta)
= ∫[0 to π/2] (1 - 3(1 - sin^2(theta)) + 3(1 - sin^2(theta))^2 - (1 - sin^2(theta))^3) d(theta).
Now, we can expand and simplify the expression further to evaluate the integral.