Evaluate the integrals in Exercises 51–56 by making a substitu...

Question

Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.

∫ (from 0 to √3/2) dy / (1 - y²)^(5/2)

Accepted Answer

Welcome back everyone. Evaluate the integral by using a trigonometric substitution. Integral from 0 to square root of 2 divided by 2 of d x divided by 1 minus x 2 traced to the power of 7 halves. A 15 divided by 28. B 28 divided by 15. C 28 divided by 5. And D 5 divided by 28. For this problem we're going to use substitution. Let's suppose that x is equal to sine theta. Now we are going to find the derivative of x with respect to theta, and that's the derivative of sine theta which is cosine theta. We can now show that d x is equal to cosine theta multiplied by d theta. And from here we want to change the limits of integration. If x is equal to 0, let's call it x1, the lower limit of integration, then theta 1 is going to be inverse of sin of 0, which is 0. For the upper limit of the integration x 2, we have square root of 2 divided by 2. So theta 2 is going to be inverse of sine of. The square root of 2 divided by 2, which is pi divided by 4. And now we can rewrite the integral as follows. The limits of integration are going to be from 0 up to pi divided by 4. Dx becomes cosine theta d theta, and in the denominator we're going to have 1 minus. Sine squared theta. Raise the power of 7 halves. And now we can perform simplifications so that be integral from 0 up to pi divided by 4 of cosine theta d theta using the Pythagorean identity. 1 minus sine squared theta is cosine squared theta, and we are raising it to the power of 7 halves. Using the properties of exponents, we can show that the the denominator is cosine of theta raised to the power of 7 halves multiplied by 2 gives us 7. And essentially we're not using the absolute value because the limits of integration are from 0 up to pi divided by 4. So, cosine is non-negative, right? Meaning we can write cosine to the power of 7 of theta. And from here we can simplify. Cosine theta can be canceled out in the numerator, and in the denominator we will have cosine to the power of 6 of theta. And this is going to be the integral from 0 to pi divided by 4 of 2 to the power of 6 theta d theta. Now from here we're going to use another substitution. Specifically, we're going to split sequence to the power of 6s. 22 theta multiplied by sequent to the power of 4 theta. Now sequence to the power of 4 can be written as sequin 2 theta squared, right? And now let's use trigonometry. So that'll be sequin 2 theta multiplied by. 1 + tian square theta squared. Well done. And now what we're going to do is simply use use substitution. Let's suppose that U is equal to tangent theta. The derivative of you with respect to the it is going to be the derivative of tangent, which is. Sequences square theta. So that 2 squared theta d theta is going to be equal to DU. Now, let's start writing the integral in terms of U and DU. So, sequence to the power of 6 is basically sequin squared. Multiplied by 1 plus tent squared. Squared, right? So what we're going to do is simply write 2 squared theta, Dheta as DU. And then 1 plus tangent squared is going to be 1 plus. U squared. So we get integral of 1 + u2 2. And the remaining part, which is s and squared, the theta is going to be the EU. Now, let's find the new limits of integration. You want the lower limit of integration. It's going to be tangents of 0, which is 0. For the upper limit of integration, U2 is going to be tangent of 5 divided by 4, which is 1. So we get integral from 0 to 1 of 1 + u2 2 d u and this is now an elementary integral. We have integral from 0 to 1, we can square 1 + u 2, which is going to give us 1 + 2U2 + u to the power of 4 DU. Let's call it integral i and continue below. We can now integrate using the power rules so that I is going to be the integral of 1 vu is simply u. Plus 2, the integral of u squared is u cubed divided by 3. Plus the integral of u to the power of 4 is u to the power of 5 divided by 5. And we went to evaluate the result between 0 and 1. At 0, every term vanishes, right, because every term contains u. So we can just evaluate the result at u equals 1. We're going to get 1 plus. 2/3 multiplied by 1 cubed, which is 2/3, plus 1 to the power of 5, which is 1 divided by 5. And now we're going to find the common denominator, which is 15. In the numerator we get. 15 Plus 5 multiplied by 2, that's 10 plus. 3 multiplied by 1 gives us 3, so overall that's 28 divided by 15. And now considering the answer choices, we can conclude that. The correct answer for this problem. Corresponds to the answer choice be. Thank you for watching.

Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.∫ (from 0 to √3/2) dy / (1 - y²)^(5/2)

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Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.
∫ (from 0 to √3/2) dy / (1 - y²)^(5/2)