How do you integrate #(sqrt(12+4x^2)dx#?
The integral can be written:
Since:
Now remembering that:
And now, remembering that:
than:
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To integrate √(12 + 4x^2) dx, you can use the trigonometric substitution method. Let x = √(3/2) * tan(θ), then dx = √(3/2) * sec^2(θ) dθ. After substitution and simplification, the integral becomes:
∫ √(12 + 4x^2) dx = ∫ √(12 + 4(√(3/2) * tan(θ))^2) * √(3/2) * sec^2(θ) dθ.
Simplify the expression inside the square root:
= ∫ √(12(1 + tan^2(θ))) * √(3/2) * sec^2(θ) dθ = ∫ √(12sec^2(θ)) * √(3/2) * sec^2(θ) dθ = ∫ √(12) * sec(θ) * √(3/2) * sec^2(θ) dθ = ∫ 2√(3) * sec^3(θ) dθ.
Now, use the integral of sec^3(θ) which is a known integral, leading to:
= 2√(3) * (1/2) * (sec(θ) * tan(θ) + ln|sec(θ) + tan(θ)|) + C = √(3) * (sec(θ) * tan(θ) + ln|sec(θ) + tan(θ)|) + C.
Finally, substitute back for θ using the original substitution x = √(3/2) * tan(θ), which implies tan(θ) = x/√(3/2) and sec(θ) = √(2/3) / √(1 - x^2/2), resulting in:
= √(3) * (√(2/3) / √(1 - x^2/2) * (x/√(3/2)) + ln|√(2/3) / √(1 - x^2/2) + (x/√(3/2))|) + C = (x√(2/3) / √(1 - x^2/2) + ln|√(2/3) / √(1 - x^2/2) + (x/√(3/2))|)√(3) + C.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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