# How do you find the integral of #5/[sqrt(9x^2-16)] dx#?

The answer is

Perform some simplification

Therefore, the integral is

Therefore,

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To find the integral of ( \frac{5}{\sqrt{9x^2-16}} ) with respect to ( x ), you can use the trigonometric substitution method. Let ( x = \frac{4}{3} \sec(\theta) ). Then ( dx = \frac{4}{3} \sec(\theta) \tan(\theta) d\theta ). Substituting these into the integral, it simplifies to ( \int \frac{20\sec(\theta)\tan(\theta)}{\sqrt{16\sec^2(\theta)}} d\theta ). This simplifies further to ( \int \frac{20\tan(\theta)}{4|\sec(\theta)|} d\theta ). Since ( |\sec(\theta)| = \sec(\theta) ) for ( \theta ) in the domain of the integral, this becomes ( \int \frac{5\tan(\theta)}{\sec(\theta)} d\theta ). Using the identity ( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} ) and ( \sec(\theta) = \frac{1}{\cos(\theta)} ), the integral becomes ( \int 5\sin(\theta) d\theta ). Integrating ( \sin(\theta) ) yields ( -5\cos(\theta) + C ), where ( C ) is the constant of integration. Finally, substituting back ( \cos(\theta) = \frac{4}{3x} ) gives the result ( \int \frac{5}{\sqrt{9x^2-16}} dx = -\frac{5}{3}\sqrt{9x^2-16} + C ), where ( C ) is the constant of integration.

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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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