85. Explain why or why not Determine whether the following sta...

Question

85. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

a. More than one integration method can be used to evaluate ∫ (1 / (1 - x²)) dx.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the validity of the given statement, and we're given the statement that there is more than one integration method that could be used to evaluate the integral of 1 divided by the quantity of X2 minus 4 and quantity DX. And we're given two possible choices as our answers. For choice A, we have true, and for choice B, we have false. So, in order to show that this statement is true, we just need to find two methods of integration that will work to evaluate our expression here. So the first method that we can try is actually partial fractions. Since 1 divided by the quantity of X2 minus 4. Can be written as being equal to 1 divided by the quantity of the quantity of X minus 2 in quantity multiplied by the quantity of X plus 2 in quantity. So here we just factored the denominator since it's a difference of 2 squares. And now we can break this single fraction into two fractions using partial fractions. So this is going to be equal to A. Divided by The quantity of X minus 2 in quantity, plus B divided by the quantity of X plus 2 in quantity. And so to make sure that we can actually um use partial fractions, we need to find the values for A and B. So, we're going to multiply our expression here by the quantity of X minus 2 in quantity multiplied by X, the quantity of X plus 2 in quantity. In doing so, we will have 1 is equal to a multiplied by the quantity of X plus 2. In quantity, plus B multiplied by the quantity of X minus 2. So to find the value for A, we'll set X equal to 2. So when X is equal to 2, we get 1 is equal to a multiplied by the quantity of 2, plus 2 in quantity, plus B multiplied by the quantity of. 2 minus 2 in quantity. And so simplify this expression, we see that 1 would be equal to 4 A. And so that means that A is going to be equal to 1 divided by 4. Now, if we set X equals to minus 2, we should be able to find the value for B. So we'll have 1 is equal to a multiplied by the quantity of -2 + 2 in quantity, plus B multiplied by the quantity of -2 minus 2 in quantity. So simplifying we'll have 1 is equal to -4 B. And B is going to be equal to minus 1 divided by 4. So now we can rewrite our original integral. So we have the integral of 1 divided by the quantity of X2 minus 4 in quantity DX, and this is going to be equal to the integral of 1 divided by the quantity of 4 multiplied by the quantity of X minus 2 in quantity, DX. Minus the integral of 1 divided by the quantity of 4 multiplied by the quantity of X plus 2 in quantity DX. And so we actually have two integrals now, and actually both of these can be integrated directly by using new substitution. So setting U equal to X minus 2 and X + 2 respectively into each of those intervals, we'll get A one divided by UDU integral to integrate, which is just our natural log um function result. So both of these integrals are going to give us natural logs, so they are integrable and therefore, this is a valid method. Or evaluating are integral. So now we just need to find one more method. And the other method that we can try is trick substitution. So we're going to set X. Equal to 2 multiplied by sequent theta, and so that means that DX is going to be equal to 2 multiplied by sequent theta, multiplied by tangent of theta D theta. And so our original integral of 1 divided by quantity of X2 minus 4 in quantity DX. is going to be equal to the interval of 2 multiplied by the. Multiplied by tangent of data. Divided by the quantity of the quantity of 2 multiplied by sequence of theta. In quantity squared -4. In quantity d theta. And so, we're going to rewrite our denominator here to make it a little bit easier to use. So, this is going to be equal to the integral of 2 multiplied by sequent of data, multiplied by tangent of data. Divided by the quantity, and here when I square the 2 in the first term in our denominator, I'll get 4. And so I can factor out a 4 out of both terms, so we'll have 4 multiplied by the quantity of squared data minus 1 in quantity, and all that is multiplied by the data. Now, we can use our trigger identities to replace sequence square theta minus 1 with tangent square theta. So, this means that our integral is going to be equal 2, and here we're gonna pull the constants out front. So, we'll have 2 divided by 4, which is 1 divided by 2, multiplied by the integral of decant data. Multiplied by tangent data. Divided by Tangent squared theta. The Theta So here we can simplify our integrand. So one factor of tangent of theta in the denominator will cancel out with the tangent theta in the numerator and recall that tangent of theta is equal to sequent of theta divided by cosequent of theta. So simplifying our integram, this is going to be equal to 1 divided by 2 multiplied by the integral of cosequent of theta D theta. And we can integrate directly cosequent of theta with respect to theta. And when we do that integration, we'll also get natural log functions as in our result. So we would get the same result either way, either by using uh partial fractions or by trig substitutions. So this here is the second valid method. That we can use to evaluate our integral. And so, since we found two methods here, that means that our statement is true, since that is more than one method. So this here is the answer that we're looking for for this problem, and we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

85. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. More than one integration method can be used to evaluate ∫ (1 / (1 - x²)) dx.

Key Concepts

Integration Methods

Improper Integrals

Trigonometric Substitution

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