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¡Ò dy/(rad£¨4-y^2))=
Evaluate the indefinite integral
look into trig substitutions
let y = 2sinu
then 4 - y^2 = 4 - 4sin^2 u = 4 cos^2 u
and √(4 - y^2) = 2cos u
dy = 2 cosu du
and your integral is just
1/(2 cosu) * 2 cosu du = 1/2 du
integral(1/2) du = u/2 = (1/2) arcsin(y/2) + C
oops - too many 1/2's there. integral(du) = u = arcsin(y/2)+C
To evaluate the given expression, we need to integrate the expression on the left-hand side of the equation with respect to y.
The integral of dy/(√(4-y^2)) can be evaluated by using a trigonometric substitution. Let's proceed with the following steps:
Step 1: Identify the trigonometric substitution.
Since the expression contains a square root term with y^2, it suggests using a trigonometric substitution involving a right triangle. In this case, we can substitute y = 2sinθ.
Step 2: Determine the limits of integration.
To replace the limits of integration, we need to find the values of y correspond to the limits. Without any given limits, we assume the limits are from y = -2 to y = 2.
When y = -2, we have sinθ = y/2 = -2/2 = -1.
When y = 2, we have sinθ = y/2 = 2/2 = 1.
Therefore, the limits of integration in terms of θ are from θ = sin^(-1)(-1) to θ = sin^(-1)(1).
Step 3: Calculate the differential for the substitution.
Now, we need to find the differential dy in terms of θ. Since y = 2sinθ, we differentiate both sides with respect to θ:
dy/dθ = 2cosθ.
Step 4: Substitute the trigonometric substitution into the integral.
With the substitution y = 2sinθ, and dy = 2cosθ dθ, we can rewrite the integral as:
∫dy/(√(4-y^2)) = ∫(2cosθ)/(√(4-(2sinθ)^2)) dθ.
Step 5: Simplify the expression inside the integral.
Simplifying the expression inside the integral, we have:
∫(2cosθ)/(√(4-4sin^2θ)) dθ.
Using the trigonometric identity sin^2θ + cos^2θ = 1, we can rewrite the simplified expression:
∫(2cosθ)/(√(4cos^2θ)) dθ.
Simplifying further, we get:
∫(2cosθ)/(2cosθ) dθ.
The cosθ terms cancel out, leaving us with:
∫dθ.
Since the integral of dθ is simply θ, the final result is:
θ + C.
Step 6: Substitute back the original variable.
Now, we need to substitute back θ with the original variable y. Remember that we initially set y = 2sinθ, so we can solve for sinθ and substitute it into the expression θ + C.
For y = 2, we have sinθ = y/2 = 2/2 = 1. This corresponds to θ = π/2.
For y = -2, we have sinθ = y/2 = -2/2 = -1. This corresponds to θ = -π/2.
Substituting these values back into θ + C, we get:
For y = 2, the result is π/2 + C.
For y = -2, the result is -π/2 + C.
Therefore, the evaluated expression is:
π/2 + C or -π/2 + C.
Note that C represents the constant of integration, which can take any value.