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.

∫ dx / (4 - x²)^(3/2) from 0 to 1

Accepted Answer

Welcome back everyone. Evaluate the integral from 0 to 1/2 of d x divided by 1 minus x 2. Raise to the power of 3 halves. We're going to approach this problem using substitution. Let's suppose that our independent variable x is equal to sine theta. And let's remember that our next step is to find the derivative of x in this case with respect to the. That's the derivative of sine theta, and the derivative of sine theta is cosine theta. So we can now express dx, which is going to be cosine theta multiplied by d theta. In other words, we're multiplying both sides of our equation by d theta. From here let's change the limits of integration in terms. Of theta and specifically we know that the lower limit of integration X1 is 0, right? So if x1 is equal to 0, then theta 1 is going to be equal to. Inverse of sin, right? We are going to show that theta is equal to inverse of sin of x if we take inverse of sin of both sides of the relationship x equals sine theta. So theta 1 is going to be inverse of sin of 0, which is 0, and the upper limit of integration is 1/2. So in terms of theta 2, that would be inverse of sin of 1/2. Which is equal to pi divided by 6. This allows us to rewrite the integral in terms of theta and the theta. We're going to integrate from 0 up to pi divided by 6. And in the numerator, dx becomes cosine theta d theta. Considering the denominator we have 1 minus x2, which is 1 minus sin 2 theta, raise to the power of 3 halves. From here, let's go ahead and simplify, so that's integral from 0 to pi divided by 6. Cosine theta d theta. Divided by using the Pythagorean identity 1 minus sine squared theta is cosine squared. Theta. And it is raised to 3 halves. Now let's use the properties of exponents. 2 multiplied by 3 halves gives us 3. So in the denominator overall we have cosine q. Theta And from here we can clearly see that we can cross out cosine theta in the numerator and in the denominator, we're going to reduce the exponent to Square So what we're actually integrating is the integral from 0 to pi divided by 6 of the theta. Divided by cosine squared theta. Now let's remember that 1 divided by cosine theta is 2. So we can also rewrite the integrals integral from 0 up to pi divided by 6 of sequin squared theta d theta, which is basically an elementary integral. And its value is tangent theta. We want to evaluate the result from 0 to pi divided by 6 using the fundamental theorem of calculus part 2. That's tantrums. Of pi divided by 6 minus tangent of 0. Since tangent of 0 is equal to 0, we can basically cross out that term. And we can focus on the remaining one, which is tangent of pi divided by 6. tangent of pi divided by 6 is the square root of 3 divided by 3, which is our final answer for this problem. Thank you for watching.

Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.∫ dx / (4 - x²)^(3/2) from 0 to 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.
∫ dx / (4 - x²)^(3/2) from 0 to 1