Evaluate the integral.
from 0 to π/2
sin^3(θ)cos^5(θ) dθ
To evaluate the integral, we'll use the power-reducing formulas to rewrite the integrand. The power-reducing formulas are:
sin^2(θ) = (1 - cos(2θ))/2
cos^2(θ) = (1 + cos(2θ))/2
Using these formulas, we can rewrite the integrand as:
sin^3(θ)cos^5(θ) = sin(θ) * sin^2(θ) * cos^4(θ) * cos(θ)
= sin(θ) * (1 - cos(2θ))/2 * (1 + cos(2θ))^2 * cos(θ)
Now, let's simplify this expression further:
sin(θ) * (1 - cos(2θ))/2 * (1 + cos(2θ))^2 * cos(θ)
= 1/2 * sin(θ) * (1 - cos^2(2θ)) * (1 + 2cos^2(2θ) + cos^4(2θ)) * cos(θ)
= 1/2 * sin(θ) * (1 - cos^2(2θ)) * (1 + 2cos^2(2θ) + cos^4(2θ)) * cos(θ)
= 1/2 * sin(θ) * (1 - (1 - sin^2(2θ))) * (1 + 2(1 - sin^2(2θ)) + (1 - sin^2(2θ))^2) * cos(θ)
= 1/2 * sin(θ) * (1 - 1 + sin^2(2θ)) * (1 + 2 - 2sin^2(2θ) + 1 - 2sin^2(2θ) + sin^4(2θ)) * cos(θ)
= 1/2 * sin(θ) * (sin^2(2θ)) * (4 - 4sin^2(2θ) + sin^4(2θ)) * cos(θ)
= 1/2 * sin^3(2θ) * (4 - 4sin^2(2θ) + sin^4(2θ)) * cos(θ)
Now, we can simplify the integral to:
∫ sin^3(θ)cos^5(θ) dθ = 1/2 * ∫ sin^3(2θ) * (4 - 4sin^2(2θ) + sin^4(2θ)) * cos(θ) dθ
At this point, the integral cannot be solved easily by hand. It may require advanced techniques such as trigonometric substitutions or numerical methods to approximate the value.
The way this is done is to express everything in terms of sin or cos:
sin^3(θ)cos^5(θ) dθ
= sin^3θ cos^4θ cosθ dθ
= sin^3θ (1-sin^2θ)^2 cosθ dθ
Now let
u=sinθ
du = cosθ dθ
and you have
u^3 (1-u^2)^2 du
and it's a cinch...