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.

∫ (1 - x²)^(1/2) / x⁴ dx

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. Evaluate the integral, the integral of the square root of 4 minus x 2, all divided by x 4 dx. OK, so it appears for this particular problem, we're asked to ultimately evaluate this integral. So now that we know that we're evaluating this integral. What we must recall and do first is we need to recall how to use trigometric substitution for X is equal to 2 multiplied by sine of theta, then that will mean. That we can go ahead and write that. DX by Dheta will be equal to 2 multiplied by cosine of theta, which will mean we can simply state that DX is equal to 2 multiplied by cosine of theta d theta. Fantastic. Thus we can go ahead and write that the square root of 4 minus x 2 will be equal to the square root of 4 minus 2 multiplied by sin of theta to the power 2, which will be equal to the square root of 4 minus 4 multiplied by sin to the 2 of theta. Which will be equal to the square root of. Or multiplied by parentheses of 1 minus. Sine to the power of 2, so sin to the power of 2 of theta, which will be equal to the square root of 4 multiplied by cosine squared of theta, which will be simply equal to 2 multiplied by cosine of theta. So now, we need to substitute back, we need to substitute our value back into our integral, and we can go ahead and write that the integral of, so once again the integral of the square root of 4 minus x 2 divided by x 4, DX will be equal to the integral of. 2 multiplied by cosine of theta. Divided by parentheses 2 multiplied by sine of theta to the power of 4 multiplied by 2 multiplied by cosine of theta d theta. And this will be equal to the integral of or multiplied by cosine squared of theta. Divided by 16 multiplied by sine to the power of 4 of theta d theta, which will be equal to. 1 divided by 4 multiplied by the integral of 1 minus sine to the power of 2 of theta divided by sine to the power of 4 of theta d theta. Which will be. Equal to 1 divided by 4 multiplied by the integral of parentheses of 1 divided by sine to the power of 4 of theta. Minus minus sine to the power of 2 of theta divided by sin to the power of 4 of theta d theta, which will be equal to 1 divided by 4 multiplied by the integral of parentheses of cosicant to the power of 4 of theta minus cosicant squared of theta. In parentheses, the theta. Fantastic. So now, our next step is we need to rewrite, so let's make a note, we need to rewrite cosecant to the 4 of theta is equal to cosecant squared. Of theta multiplied by cosequant squared of theta, which is equal to parentheses of 1 plus cotangent squared of theta multiplied by cosicant square of theta, which is equal to cosicant square of theta plus cotangent squared. Of theta multiplied by Cosicant squared of theta. Fantastic. So therefore we can go ahead and write. Keeping this in mind, we can go ahead and write that 1 divided by 4 multiplied by the integral of parentheses of cosicant to the power of 4 of theta minus cosicant square of theta. D theta will be equal to 1 divided by 4 multiplied by the integral of parentheses of cosequant squared of theta plus cotangent squared of theta multiplied by cosequant squared of theta. Minus Cosi square of theta. D theta, which will be all equal to 1 divided by 4 multiplied by the integral of cotangent squared of theta multiplied by cosequant squared of theta d theta. Fantastic. So now our next step is we need to recall and use the substitution method and let us state that you will be equal to cotangent of theta. Then that will mean we can write that DU by D theta will be equal to negative cosequant square of theta. Which will mean that we could go ahead and write that negative DU will be equal to Cosi 2 of theta D theta. Fantastic. So we can go ahead and write 1 divided by 4 multiplied by the integral of cotangent square of theta multiplied by cosequent square of theta. D theta is equal to 1 divided by 4 multiplied by the integral of. U to the power of 2 multiplied by DU, which will be equal to 1 divided by 4 multiplied by the integral of u to the 2DU. Fantastic. So, moving right along here, we now need to recall and use the power rule, which states that the integral of U to the power of NDU. Is equal to drumroll, U to the power of n + 1 divided by n + 1 + C, and this is for where N cannot equal -1. Once again, N, this integer n cannot equal -1. So keeping that in mind and applying the power rule, we could go ahead and write -1 divided by 4 multiplied by the integral of u to the 2Du is equal to 1 divided by 4 multiplied by parentheses of u to the 3 divided by 3 plus c, where this plus c is some constant value. And this will be equal to -1 divided by 12 multiplied by U to the power of 3 plus C. So now our next step is we need to back substitute U is equal to cotangent of theta. So when we back substitute, we will be able to write -1 divided by 12 multiplied by cotangent to the power of 3. Of data Plus C, fantastic. And we need to note that since X is equal to 2 multiplied by sin theta, that will mean that sin theta is equal to x divided by 2. And that will mean that cosine theta is equal to the square root of 1 minus sin square of theta. Which will be equal to square root of 1 minus x divided by 2 to 2, which will be equal to the square root of 1 minus x 2 divided by 4, which will be equal to the square root of 4 minus x 2 divided by 4. Which will be equal to the square root of 4 minus X 2 divided by 2. Thus, that will mean that cotangent of theta will be equal to cosine of theta divided by sin theta, which will be equal to the square root of 4 minus x 2 divided by 2, all divided by X divided by 2. Which will be simply equal to the square root of 4 minus x 2 divided by X. So therefore, without a doubt, we could go ahead and write that -1 divided by 12 multiplied by cotangent to the power of 3 of theta plus c will be equal to -1 divided by 12 multiplied by parentheses of the square root of 4 minus x to the power of 2 divided by. X All to the power of 3 plus c and this is our correct and final answer. Hooray, we did it. So going back to the top to recap what we've learned and to quickly restate our final answer, we will determine that without a doubt our final answer. For the evaluated integral is once again. 1 divided by 12 multiplied by parentheses of the square root of 4 minus x 2, all divided by X 3 plus C. 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.∫ (1 - x²)^(1/2) / x⁴ dx

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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.
∫ (1 - x²)^(1/2) / x⁴ dx