How do you integrate #int 1/sqrt(x^2-16x+3) # using trigonometric substitution?
NOTE : -It is better to use
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Use the substitution
Let
Complete the square in the square root:
Simplify:
Integrate directly:
Reverse the substitution:
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To integrate ( \int \frac{1}{\sqrt{x^2 - 16x + 3}} ) using trigonometric substitution, let ( x = 8 + \sqrt{3} \sec(\theta) ). Then differentiate ( x ) with respect to ( \theta ) and substitute into the integral. This leads to ( dx = \sqrt{3} \sec(\theta) \tan(\theta) d\theta ). Substitute ( x ) and ( dx ) into the integral, simplify, and then use trigonometric identities to express the integrand in terms of trigonometric functions. Finally, evaluate the integral using trigonometric identities and simplify the 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.
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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