finf the definite integral of the following h(u)=cos^2(1/8*u) (8th*u)
To find the definite integral of the function h(u) = cos²(1/8u) * (8u), we can use integration techniques. The process involves a combination of trigonometric identities, substitution, and integration rules.
1. Start by simplifying the function:
h(u) = cos²(1/8u) * (8u)
= (cos²(1/8u))(8u)
= 8u * cos²(1/8u)
2. Apply the double-angle identity for cosine:
cos(2θ) = 2cos²(θ) - 1
Notice that our function closely resembles the form of the double-angle identity.
Let θ = 1/8u and substitute:
cos(2θ) = 2cos²(θ) - 1
cos(2(1/8u)) = 2cos²(1/8u) - 1
Rearrange the equation to isolate cos²(1/8u):
cos²(1/8u) = (1/2)cos(2(1/8u)) + 1/2
Substitute cos²(1/8u) back into our original function:
h(u) = 8u * [(1/2)cos(2(1/8u)) + 1/2]
= 4u * cos(2(1/8u)) + 4u
3. Integration:
Now, we integrate the function h(u) with respect to u. We will perform each integration term-by-term.
∫[4u * cos(2(1/8u)) + 4u] du
= ∫[4u * cos(2(1/8u))] du + ∫[4u] du
Integration is linear, so we can separately integrate each term.
∫[4u * cos(2(1/8u))] du + ∫[4u] du
= 4 * ∫[u * cos(2(1/8u))] du + 4 * ∫[u] du
The first integral, ∫[u * cos(2(1/8u))] du, requires substitution.
Let w = 2(1/8u), which implies du = 8dw.
The limits of integration will also change.
When u = a, w = 2(1/8a)
When u = b, w = 2(1/8b)
Substituting, we get:
4 * ∫[u * cos(w)] * (1/8) dw + 4 * ∫[u] du
Simplifying, we have:
(1/2) * ∫[u * cos(w)] dw + 4u²
Applying integration rules:
(1/2) * ∫[u * cos(w)] dw = (1/2) * u * sin(w) - (1/2) * ∫[sin(w)] du
The integral of sin(w) is a straightforward integration:
∫[sin(w)] dw = -cos(w)
Plugging the result back in, we have:
(1/2) * u * sin(w) - (1/2) * (-cos(w))
Leaving us with:
(1/2) * u * sin(w) + (1/2) * cos(w) + 4u²
4. Substitute the original angle value:
Since we initially set w = 2(1/8u), substitute back to get:
(1/2) * u * sin(2(1/8u)) + (1/2) * cos(2(1/8u)) + 4u²
5. Evaluate the definite integral:
To find the definite integral, apply the upper and lower limits of integration.
The definite integral of h(u) from a to b is:
∫[a to b] (1/2) * u * sin(2(1/8u)) + (1/2) * cos(2(1/8u)) + 4u² du
Evaluate this expression by substituting the upper bound (b) and lower bound (a) into the integral.