# How do you integrate #int 1/sqrt(t^2-6t+13)# by trigonometric substitution?

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To integrate ( \int \frac{1}{\sqrt{t^2 - 6t + 13}} ) using trigonometric substitution:

- Complete the square inside the square root: ( t^2 - 6t + 13 = (t - 3)^2 + 4 ).
- Recognize that this resembles the form ( a^2 + x^2 ), where ( a = 2 ).
- Substitute ( t - 3 = 2 \sin(\theta) ), which implies ( t = 2\sin(\theta) + 3 ), and ( dt = 2\cos(\theta) d\theta ).
- Rewrite the integral in terms of ( \theta ): ( \int \frac{1}{\sqrt{(2\sin(\theta) + 3)^2 + 4}} \cdot 2\cos(\theta) d\theta ).
- Simplify the integrand using the substitution.
- Integrate the simplified expression.
- Substitute back ( \theta ) in terms of ( t ) to obtain the final result.

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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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