How do you evaluate #int1/sqrt(1-9x^2)dx# from [0,1/6]?
I have solved this way:
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To evaluate ( \int_{0}^{\frac{1}{6}} \frac{1}{\sqrt{1-9x^2}} , dx ), you can use trigonometric substitution. Let ( x = \frac{1}{3}\sin(\theta) ), then ( dx = \frac{1}{3}\cos(\theta) , d\theta ). Substituting these into the integral and using the trigonometric identity ( \sin^2(\theta) + \cos^2(\theta) = 1 ), you can simplify the integral in terms of ( \theta ). After integrating, evaluate the result from ( 0 ) to ( \frac{\pi}{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.
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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