# How do you integrate #int dx/sqrt(81-9x^2)# using trig substitutions?

Let x= 3sin u, so that dx= 3 cos u du

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To integrate ( \int \frac{dx}{\sqrt{81 - 9x^2}} ) using trigonometric substitution, let ( x = \frac{9}{3}\sin(\theta) ). Then, ( dx = \frac{9}{3}\cos(\theta) d\theta ). After substitution and simplification, the integral becomes ( \int \frac{9\cos(\theta)d\theta}{\sqrt{81 - 81\sin^2(\theta)}} ). This simplifies to ( \int \frac{9\cos(\theta)d\theta}{9\cos(\theta)} ), which reduces to ( \int d\theta ). Hence, the integral becomes ( \theta + C ), where ( C ) is the constant of integration. Finally, substitute back ( x = \frac{9}{3}\sin(\theta) ) to get the answer in terms of ( x ).

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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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