Evaluate the integrals in Exercises 67–74 in terms of
b....

Question

Evaluate the integrals in Exercises 67–74 in terms of

b. natural logarithms.

73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)

Accepted Answer

Welcome back everyone. Evaluate the integral from 0 to pi divided by 2 of sin x d x divided by square root of 1 plus cosine squared x. A ln of square root of 2 minus 1. B 1 divided by ln of 1 + square root of 2. Cln of 1 + square root of 2. In the 1 divided by ln of square root of 2 minus 1. For this problem, we're going to use to use substitution. Let's suppose that U is equal to cosine x. The next step is to find the derivative of U with respect to X, and that's the derivative of cosine, which is negative sin x. If we multiply both sides by negative dx, we can show that negative DU is going to be equal to sin x dx, and this is exactly what we have within the integral. So what we want to do is change the elements of integration. The lower one is going to be cosine of 0, which is 1, and the upper one is going to be cosine of pi divided by 2, which is going to be equal to 0. So now we are ready to rewrite the integral in terms of U and DU. We're integrating from 1 up to 0. Sin XDX becomes negative DU, we can factor out the negative sign. And in the denominator we have the square root of 1 + u 2. We can apply the properties of integrals. Specifically, we can reverse. The limits, we can integrate from 0 to 1 instead, which basically Means that we're adding an additional negative sign in front and negative of negative means that we get the positive sign now. Well done, and this is an elementary integral. It is equal to inverse of singe of u. Evaluate that from 0 to 1. But since the answers are provided in terms of natural logarithms, we can just recall that this is L and of the absolute value of U plus square root of 1 + u 2, right? Evaluated from 0 to 1. So let's go ahead and substitute the values. At the upper limit we get ln of 1 + square root of 1 + 12, and that's the square root of 2, right, minus ln of the absolute value of 0 plus square root of 1 + 02. So that's basically the square root of 1, which is 1, right? And L of 1 is equal to 0. So we can just cross out that term because it is equal to 0, and we can drop the absolute value for 1 + square root of 2 because it is positive. This is how we get our final answer. It is LN of 1 + square root of 2, and this corresponds to the answer choice C. Thank you for watching.

Evaluate the integrals in Exercises 67–74 in terms ofb. natural logarithms.73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)

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Evaluate the integrals in Exercises 67–74 in terms of
b. natural logarithms.
73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)