evaluate the following integral using trigonometric substitution.
integral (dx)/(square root 289-x^2))
Rewrite given interval using this substitution.
Evaluate the integral.
To evaluate the integral
∫dx/√(289-x^2),
we can use the trigonometric substitution x = 17sin(θ), where -π/2 ≤ θ ≤ π/2.
When x = 17sin(θ), we can rewrite the expression inside the square root as follows:
289 - x^2 = 289 - (17sin(θ))^2 = 289 - 289sin^2(θ) = 289(1 - sin^2(θ)) = 289cos^2(θ).
From this, we can see that dx = 17cos(θ)dθ.
Substituting x = 17sin(θ) and dx = 17cos(θ)dθ into the integral, we get:
∫dx/√(289-x^2) = ∫17cos(θ)/√(289cos^2(θ)) dθ.
Simplifying the expression further:
∫17cos(θ)/17cos(θ) dθ = ∫dθ.
The integral of a constant term is simply the variable itself, so we have:
∫dθ = θ + C.
Finally, we need to convert back from θ to x. Using the original substitution x = 17sin(θ), we solve for θ:
x = 17sin(θ) ⟹ θ = arcsin(x/17).
Thus, the final answer is:
∫dx/√(289-x^2) = arcsin(x/17) + C.