Evaluate the integrals in Exercises 67–74 in terms of a. inverse hyperbolic functions. 73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)

Welcome back everyone. Evaluate the following definite integral integral from 0 to 2 pi of sin x d x divided by square root of 1 plus cosine squared x. A 0 B2 inverse of cinch of 1, C inverse of cinch of 2, and d inverse of cash of 2. For this problem, we're going to use use substitution. Let's suppose that u is equal to cosine x. Then the derivative of PU with respect to x is going to be. The derivative of cosine x, which is negative sine x. In the numerator we have sin x dx, so let's rearrange this expression for sinx multiplied by dx. Which is equal to DEU divided by -1 or DEU. Our next step is to change the limits of integration. The lower 1, u1, is going to be cosine of 0, which is 1, and the upper 1, U2 is going to be cosine of 25, which is also 1. So now the integral becomes integral from 1 to 1 of sin x dx becomes negative DU. In the denominator we have the square root of. 1 + U 2 And this integral is equal to 0 because the lower limit of integration is equal to the upper limit of integration, so the correct answer is option A 0. Thank you for watching.

Evaluate the integrals in Exercises 67–74 in terms ofa. inverse hyperbolic functions.73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)

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Evaluate the integrals in Exercises 67–74 in terms of
a. inverse hyperbolic functions.
73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)