Use any method to evaluate the integrals in Exercises 15–38. M...

Question

Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.

∫ (x² dx) / (x² - 1)^(5/2), where x > 1

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the integral, the integral of x 2 multiplied by dx divided by parentheses x 2 minus 9 to the power of 5 divided by 2, where x is greater than 3. OK. So it appears for this particular problem, we're asked to solve for the integral, we're asked to find the integral that is provided to us by the prom itself, where X is greater than 3. So with that in mind, now that we know what we're ultimately solving for, let us begin to solve this problem by letting i be equal to the integral of x 2 multiplied by dx divided by parentheses x 2 minus 9 to the power of 5 divided by 2. Fantastic. And let us also let X be equal to 3 multiplied by c of theta, then that will mean that we can write that DX by Dheta will be equal to 3 multiplied by sec of theta multiplied by tangent. Of theta, which will mean that dx will be equal to 3 multiplied by c cant of theta multiplied by tangent of theta d theta. Fantastic. So With this in mind, let us substitute our known variables now back into our integral. So I will now be equal to, so once again, I will now be equal to the integral of parentheses of. 3 multiplied by sec of theta to the power of 2 multiplied by 3 multiplied by sec of theta multiplied by tangent of theta d theta. All divided by parentheses of parentheses of 3 multiplied by c can of theta to the power of 2. -9, all to the power of 5 divided by 2. Fantastic. So with that said, We now need to do a little bit. We need to start solving for our integral by simplifying as we go. So, we can go ahead and write that I will be equal to the integral of 9 multiplied by C2. Of theta multiplied by 3 multiplied by C canta theta multiplied by tangina theta D theta. All divided by parentheses of 9 multiplied by c2 theta minus 9, all to the power of 5 divided by 2. So 5 divided by 2. OK, we need to keep. Simplifying until we've simplified down to our most simple form. So we need to keep going. So I is equal to the integral of 27 multiplied by c can to the power of 3 of theta multiplied by tangent. Of theta, D theta. And we need to divide this by, so divided by drum roll 9 multiplied by parentheses of C square to the minus 1. All to the power of 5 divided by 2. OK. And I will now be equal to the integral of 27 multiplied by c to the 3 of theta multiplied by tangent of theta d theta, all divided by parentheses 9 multiplied by tangent squared of theta to the power of 5 divided by 2. And so we make sure that we know that this is a power. We'll ride it a little bit. We'll ride it diagonally, but it's to the power of 5 divided by 2. So, moving right along here. And take it a step further and write that I is equal to the integral of 27 multiplied by c to the power of 3 of theta. Multiplied by tangent of theta d theta, all divided by 243 multiplied by tangent to the power of 5 of theta. And we can go even further and write that I is equal to 1 divided by 9 multiplied by the integral of C can't. To the power of 3 of theta divided by tangent to the power of 4 of theta d theta. Fantastic. We still need to keep simplifying and solving. So i is equal to 1 divided by 9 multiplied by the integral of 1 divided by cosine to the power of 3 of theta divided by sine to the power of 4 of theta divided by cosine to the power of 4 of theta. The theta. And we can go ahead and write that I is equal to 1 divided by 9 multiplied by the integral of cosine theta divided by sin to the power of 4 theta d theta. Which we could simplify. And once more, to write that i is equal to 1 divided by 9 multiplied by the integral of sin to the power of inverse. So, it's gonna be inverse sin to the power of 4. So, inverse sin to the power of 4 of theta multiplied by cosine of theta d theta. And now we need to use, rather recall and use the substitution method. So let us state that you will be equal to sin of theta. Then that will mean that DU by D theta will be equal to cosine theta, which will mean that DU is equal to cosine theta. D Theta. Fantastic. So, we can go ahead and write using our substitution method that I is equal to 1 divided by 9 multiplied by the integral of u to the power of -4. Do you And now, at this stage, we need to recall and use the power rule, and the power rule states that the integral of u to the power of NDU will be equal to u to the power of n + 1 divided by n + 1. Plus C and this will be 4 N cannot equal -1. So, once again, the value of N cannot equal 1. So with that in mind, using our power rule, we could go ahead and write that i is equal to 1 divided by 9 multiplied by the integral of u to the power of -4 DU, which will be equal to 1 divided by 9 multiplied by parentheses of u to the power of -3. Divided by -3. And then we need to add plus C. And this will be equal to -1 divided by 27 multiplied by U to the power of -3. Plus C. And now we need to back substitute, and let's note, we need to back substitute U is equal to sin of theta, and when we do, we won't be able to write that I is equal to -1 divided by 27. Multiplied by inverse sine to the power of 3 of theta plus c. And we need to make a note. We need to note that X is equal to 3 multiplied by sec of theta, which will mean that sec of theta is going to be equal to x divided by 3. So that will mean that cosine of theta will be equal to 3 divided by x. So therefore, We can make another note. Therefore, we could go ahead and write that sine of theta will be equal to the square root of 1 minus cosine to the power 2 of theta, which will be equal to the square root of 1 minus 3 divided by x to the power of 2. Which will be equal to the square root of 1 minus 9 divided by X to the power of 2. Which will be equal to the square root of x 2. -9, so it's gonna be X 2 minus 9 divided by X 2. Which will be equal to the square root of x 2 minus 9 divided by x. Fantastic. So therefore, with this in mind, we can go ahead and write that I is equal to negative. 1 divided by 27 multiplied by parentheses of the square root of x 2 minus 9 divided by x to the power of -3 plus c. Which will be equal to -1 divided by 27 multiplied by parentheses of x divided by the square root of x to the 2 minus 9. To the power of 3 plus c and this will be our correct and final answer. Hooray, we did it. So going back to the top to recap what we've learned, we can safely say that our final answer for the integral will be once again. Drum roll, 20, so it's gonna be -1 divided by 27 multiplied by parentheses of X divided by the square root of X2 minus 93 plus C. And that's it. That is our correct and final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.∫ (x² dx) / (x² - 1)^(5/2), where x > 1

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Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.
∫ (x² dx) / (x² - 1)^(5/2), where x > 1