How do you integrate #int 1/sqrt(x^29x7) # using trigonometric substitution?
For this, you're going to have to complete the square.
Now bring it back into the problem.
What we now have is:
Remember the trick to do this? See the following:
So the answer is:
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To integrate ( \int \frac{1}{\sqrt{x^2  9x  7}} ) using trigonometric substitution, follow these steps:

Complete the square inside the square root. [ x^2  9x  7 = (x^2  9x + \frac{81}{4})  \frac{81}{4}  7 = (x  \frac{9}{2})^2  \frac{121}{4} ]

Make the substitution ( x  \frac{9}{2} = \frac{11}{2} \sec \theta ), which implies ( x = \frac{11}{2} \sec \theta + \frac{9}{2} ), and ( dx = \frac{11}{2} \sec \theta \tan \theta , d\theta ).

Substitute ( x ) and ( dx ) in terms of ( \theta ) and simplify the expression: [ \frac{1}{\sqrt{x^2  9x  7}} = \frac{1}{\sqrt{(\frac{11}{2} \sec \theta + \frac{9}{2})^2  \frac{121}{4}}} ]

Simplify the expression inside the square root.

Make the trigonometric substitution and integrate the expression in terms of ( \theta ).

Substitute back ( x ) in terms of ( \theta ) to find the final result.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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