How do you integrate #int sqrt(54xx^2) dx# using trigonometric substitution?
We have:
The first integral can be approached with substitution, or just the reverse chain rule. The second is simple:
Writing all in terms of sine:
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To integrate ∫√(54xx²) dx using trigonometric substitution, follow these steps:

Complete the square inside the square root: 5  4x  x² =  (x² + 4x  5) =  [(x² + 4x + 4)  9] =  [(x + 2)²  9] =  (x + 2)² + 9

Substitute x + 2 = 3sin(θ), which implies dx = 3cos(θ) dθ

After substitution, the integral becomes: ∫√(9cos²(θ)) * 3cos(θ) dθ

Simplify the expression inside the square root: √(9cos²(θ)) = 3cos(θ)

Replace the integral in terms of θ: 3∫cos(θ) * 3cos(θ) dθ

Break the integral into two parts based on the domain of cos(θ): 0 to π/2 and π/2 to π

Integrate each part separately.

After integration, substitute back θ with its original expression involving x.

Simplify the expression to obtain the final result.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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