# How do you integrate #int sqrt(-x^2-6x+16)/xdx# using trigonometric substitution?

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To integrate ( \int \frac{\sqrt{-x^2-6x+16}}{x} , dx ) using trigonometric substitution, let ( x = 4\sin(\theta) ). Then ( dx = 4\cos(\theta) , d\theta ). Substituting these into the integral, we get ( \int \frac{\sqrt{-(16\sin^2(\theta) + 6(4\sin(\theta)) + 16)}}{4\sin(\theta)} \cdot 4\cos(\theta) , d\theta ). Simplifying the expression under the square root gives ( \sqrt{-16\sin^2(\theta) - 24\sin(\theta) + 16} = \sqrt{16(\sin^2(\theta) + 1.5\sin(\theta) - 1)} ). This can be further simplified using trigonometric identities. Let ( u = \sin(\theta) + \frac{3}{4} ). Then ( du = \cos(\theta) , d\theta ). Substituting these into the integral and simplifying yields ( \int \sqrt{16(u^2 - \frac{9}{16})} , du ). This simplifies to ( \int 4\sqrt{u^2 - \frac{9}{16}} , du ). Finally, integrate using standard techniques for integrals of the form ( \int \sqrt{u^2 - a^2} , du ) where ( a ) is a constant.

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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.

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