76. Apparent discrepancy
Three different computer algebr...

Question

76. Apparent discrepancy

Three different computer algebra systems give the following results:

∫ (dx / (x√(x⁴ − 1))) = ½ cos⁻¹(√(x⁻⁴)) = ½ cos⁻¹(x⁻²) = ½ tan⁻¹(√(x⁴ − 1)).

Explain how all three can be correct.

Accepted Answer

Hello. In this video, we are going to be verifying which of the following options is equivalent to the given indefinite integral. Now, the indefinite integral given to us is the integral of 1 divided by T multiplied by the square root, a T4th power minus 9 DT. Now, we can solve this integral by using trigonometric substitution. The square root function is in the form of X2 minus A2, which means we can use a trigonometric substitution in the form X is equal to A multiplied by sequent of theta. Now, in this case here, we need to identify our X squared and our A term, a squared term. Now, X2d in this case, is defined as T 4th power. And since we need X for the trigonometric substitution, by taking the square root of both sides of this equation, we get X's equal to T2d. Next for our A square term, A squared is defined as 9 in this case, and by taking the square root, we get A is equal to 3. So for the trigonometric substitution, we can allow X is equal to a sin theta, which is going to be T2 is equal to 3 multiplied by sequent of theta. Now, the thing is, if we apply a trigonometric substitution, we still need to come up with a substitution for the T in front of the square root. We can find that substitution by taking the derivative or by solving for T in the substitution, and that will give us T is equal to 3 sequent, the square root of 3 sequant of theta. And now we need to come up with a substitution of T in terms of theta. In order to do this, we will go ahead and take the derivative of both sides of this equation, and that is going to give us DT. is equal to the square root of 3 divided by 2 multiplied by the square root of C and theta. Multiplied by sequent of theta tangent of theta d theta. And by further simplifying, we can rewrite DT to equal to tangent of theta multiplied by the square root of 3 sequent of theta divided by 2 multiplied by d theta, and this is going to be the substitution of for DT in terms of theta. So, now by applying the substitutions, we can rewrite the definite integral to be the integral. Of 1 divided by T, which is now square root of 3 sequent theta. Multiplied by these square roots. Of T to the 4th power which is going to simplify to be 9 squared. -9, and this will be multiplied by the substitution for DT which is tangent of theta. Multiplied by the square root of 3 sequent of theta. Divided by 2. Multiplied by the theta. Now let's go ahead and simplify. First, notice that we have a square root, 3 and theta in both the numerator and denominator. We can simplify these out. Furthermore, we can factor out a 9 from seek and squared theta, and we will be left with a square root. Or 9, the square root of 9, multiplied by equin square theta. -1 This will further simplify to be 3. Multiplied by the square root. Of seek and square minus 1, which can be rewritten as tangent square data, which will further simplify to be 3 tangent of theta. So, we can further simplify this to be the integral of tangent of theta. Divided by 3 tangent of theta. Multiplied by 2 D theta. We have tangent in both the numerator and denominator, which will simplify out, and 3 multiplied by 2 is going to give us 6. So we can further simplify this integral to be 1/6 multiplied by the integral of D theta, and the antiderivative of D theta will just be theta. So we are left with 1/6 theta, and since we were working with an indefinite integral, we will add a general constant C. Now what we need to do is we need to convert theta back into terms of T. And we can do that by using our substitution for T2 equal to 3 sequence of theta. No Again, T2 is equal to 3 sequent of theta. If we were to solve for theta in this case, we can rewrite this to be T2d divided by 3 is equal to skin of theta, and by taking seekin inverse of both sides of the equation, we get theta is equal to seek and inverse of T2 divided by 3. And so that means we can rewrite this to be 1/6 multiplied by sequent inverse of T2 divided by 3 plus C. Now, does this, is this one of our options from A to B or C? Well, this is going to be the option for C, so at least this verifies that option C is one of the solutions. But now let's go ahead and see if we can obtain any of the other solutions. If we go back to the trigonometric identity that we solve for, we have T2 is equal to 3 sequin of theta. Now, if we were to make a right triangle in this case, With theta, recall that sequent is the inverse of cosine. And so if cosine is opposite over hypotenuse, then second is hypotenuse. I'm sorry, if cosine is adjacent over hypotenuse, then C is hypotenuse over the adjacent side to theta on the right triangle. So that is going to be T2d on the hypotenuse, and 3 on the adjacent side, and that is going to give us an opposite side to the angle theta of T 4th power minus 9. Now, because cosine is the inverse to sequent, we can also rewrite sequent inverse to be theta is equal to cosine inverse of 3 divided by T squared. And so what this means is that another solution that we can obtain is going to be theta. is equal to 1/6 multiplied by cosine inverse of 3 divided by T2 plus C, which is one of the other solutions that we have, that is equivalent to option A. Now, can we obtain option B? Well, in this case here, if we were to solve for tangent in the right triangle, we get tangent of theta is equal to the square root of T4th power minus 9 divided by 3. And if we were to solve for theta, we get theta is equal to tangent inverse. Of the square root of T4th power minus 9 divided by 3. And if we were to plug this into a solution, we get another equivalent solution where we get 1/6. Multiplied by tangent inverse. Of square root, T 4th minus 9 divided by 3 plus C, and that is going to match option B. And so what this means is that all three options are equivalent solutions or equivalent expressions to the given definite integral, and that means the overall solution to this problem is going to be D, all of the above. So I hope this video helps you in understanding how to approach this integral using trigonometric substitution, and we will go ahead and see you all in the next video.

76. Apparent discrepancy
Three different computer algebra systems give the following results:
∫ (dx / (x√(x⁴ − 1))) = ½ cos⁻¹(√(x⁻⁴)) = ½ cos⁻¹(x⁻²) = ½ tan⁻¹(√(x⁴ − 1)).
Explain how all three can be correct.

Key Concepts

Integration and Antiderivatives

Inverse Trigonometric Functions

Equivalence of Expressions

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