# Using trigonometric substitution what is the integral of #intdx/(x^2sqrt(x^2-1)# ?

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#intdx/(x^2sqrt(x^2-1)#

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To integrate ( \int \frac{dx}{x^2\sqrt{x^2-1}} ) using trigonometric substitution, we let ( x = \sec(\theta) ), then ( dx = \sec(\theta)\tan(\theta) , d\theta ). Substituting these expressions into the integral and simplifying yields:

[ \int \frac{dx}{x^2\sqrt{x^2-1}} = \int \frac{\sec(\theta)\tan(\theta)}{\sec^2(\theta)\sqrt{\sec^2(\theta)-1}} , d\theta ]

[ = \int \frac{\tan(\theta)}{\sec(\theta)\tan(\theta)} , d\theta ]

[ = \int \cos(\theta) , d\theta ]

Now, integrating ( \cos(\theta) ) with respect to ( \theta ) gives:

[ = \sin(\theta) + C ]

Since we initially let ( x = \sec(\theta) ), we need to express the result in terms of ( x ). Recall that ( \sec(\theta) = x ), so we have ( \sin(\theta) = \sqrt{x^2-1} ). Thus, the final result is:

[ \int \frac{dx}{x^2\sqrt{x^2-1}} = \sqrt{x^2-1} + C ]

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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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