# How do you find the integral of #int 1/(t^2-9)^(1/2)dt# from 4 to 6?

# int_4^6 \ 1/sqrt(t^2-9) \ dt = ln ( ( 6+sqrt(27))/(4+sqrt(7)) ) ~~ 0.5215924... #

We seek:

We are able to substitute a trigonometric function, Let

And when we put the substitution back, we obtain:

Next, we go back to the integral that is definite:

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As we are aware,

Here,

graph{1/sqrt(x^2-9) [-0.7, 4.3, 6.23, -3.77]}

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To find the integral of ( \int \frac{1}{\sqrt{t^2 - 9}} , dt ) from 4 to 6, you can use a trigonometric substitution. Let ( t = 3 \sec(\theta) ). Then ( dt = 3 \sec(\theta) \tan(\theta) , d\theta ). Substitute these values into the integral and integrate with respect to ( \theta ). Once you've found the antiderivative in terms of ( \theta ), you can convert back to the variable ( t ) to evaluate the integral from 4 to 6.

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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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