7-56. Trigonometric substitutions Evaluate the following integ...

Question

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.

12. ∫[1/2 to 1] √(1 - x²)/x² dx

Accepted Answer

Welcome back, everyone. Determine the following integral using trigonometric substitution integral from 0 to square root of 3 divided by 2 of XDX divided by square root of 1 minus X2. For this problem, let's introduce a substitution X equals sin theta. What we're going to do from here is simply identify the derivative of X with respect to theta. D X divided by the theta is the derivative of sine which is cosine of theta. So now we can show that the X is equal to cosine theta. The Sara Now, considering our integral, we have X multiplied by the X and the numerator, so X multiplied by DX is going to be equal to sine theta multiplied by cosine theta D theta. Our denominator involves square root of 1 minus X2, which is square root of 1 minus sine squared of theta, and according to the Pythagorean identity, we are going to have square root of cosine squared. Of Tegra And this is equal to cosine of data, right? We are only taking the positive solution, even though we have two potential solutions, plus or minus cosine theta. For this integration, we are going to take the positive solution, since our X is between 0 and square root of 3 divided by 2, right? So, we now have our expressions in terms of data. Our final step is to identify the new limits of integration. So what we're going to do is simply solve for theta. We know that the lower limit of integration is 0, so we're going to set 0 equals sign of theta 1, and we can essentially show that Theta 1 equals 0, right? Then we are going to solve for the upper limit of integration square root of 3 divided by 2 equals sign of theta 2, and then our upper limit of integration theta 2 is going to be equal to pi divided by. Wait And now let's go ahead and rewrite our integral, starting with the lower limit of 0, the upper limit of 5 divided by 3. Our numerator contains X multiplied by DX, which is sine theta, cosine theta D theta. And for the denominator, we have cosine of theta. We can clearly see that we can cancel out cosine of theta, and we are simply integrating sine theta D theta. The integral of sine is negative cosine of theta. And now we want to evaluate the results from 0 to pi divided by 3. Applying the fundamental theorem of calculus part 2, we take negative cosine of pi divided by 3. And we subtract negative cosine. At 0. This is equal to. 15. Plus now cosine of 0 is 1, so we're adding 1. 1/2 plus 1 is positive 1/2, meaning the answer to this problem is 1/2. Thank you for watching.

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.12. ∫[1/2 to 1] √(1 - x²)/x² dx

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7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
12. ∫[1/2 to 1] √(1 - x²)/x² dx