Definite integral sqrt(1-cos(t)) on 0 to 2pi. Two ways to integrate sqrt(1-cos(t)) and sign issue!
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Definite integral sqrt(1-cos(t)) on 0 to 2pi. Two ways to integrate sqrt(1-cos(t)) and sign issue!
00:00 Introduction to the problem: integrate sqrt(1-cos(t))dt on the interval 0 to 2pi. Preview that the integral evaluates to zero if we aren't extremely careful about the sign issues here.
00:46 Use a conjugate to integrate sqrt(1-cos(t)). The idea here is that we use the conjugate in order to build a difference of two squares inside the square root. The goal is to use the conjugate to get 1-(cos(x))^2 in the square root, which then simplifies to (sin(x))^2 inside the square root. Unfortunately, there is a sign issue lurking in our integral, and on our first attempt, it integrates to zero. This can't be correct, because the integrand is always non-negative and our integral must be positive as a result!
02:32 We identify the sign issue in the problem: the square root of the square of the sine function can't just be replaced with (sin(x))! The issue here is that our integration interval includes some values of the sine function that are positive and some that are negative. The square root of the square of sin(x) must always be positive; i.e., it's actually the absolute value of sin(x). So, we need to split our integration interval at t=pi and set up two integrals. In the second integral, we have a -sin(x) in the numerator in order to make it positive. Now we can finish the integral and get a sensible result: 4sqrt(2).
05:22 Alternate solution to the integral using the trig identity 1-cos(t)=2*(sin(t/2))^2. In this approach, we use the half-angle/double-angle identities to simplify 1-cos(t) in one step. When the integral is set up, we no longer have the sign issue that popped up in the conjugate method. This is simply because the largest argument of the sine function is now pi, which means it never becomes negative. So, we get away with ignoring the absolute value bars. We compute the integral using the chain rule backwards, and the value of the integral agrees with the first approach.
