Evaluate the integrals in Exercises 39–56.
43. ∫(from 0 ...

Question

Evaluate the integrals in Exercises 39–56.

43. ∫(from 0 to π)(sin t)/(2 - cos t) dt

Accepted Answer

Welcome back everyone. Compute the integral from 0 to pi of sin t divided by 5 minus cosine t dt. Aeon of 5/3, b eon of 3 halves, celon of 6/5s, and delon of 2/3. For this problem we're going to use to use substitution. Let's suppose that u is equal to 5 minus cosinec, and let's remember that the next step is to find the derivative of u with respect to c. That's the derivative of 5 minus cosine c. The derivative of 5 is 0. The derivative of cosinet is negative sin t. So we get negative of negative sin t, which is Sign T Multiplying both sides by DC, we can now show that DU is equal to sin CDC, which is exactly what we have within the integran, right? So now let's change the limits of integration. The lower one is going to be 5 minus cosine of 0. We know that cosine of 0 is equal to 1, so we get 5 minus 1, which is 4. And the upper one, the u2 is 5 minus cosine of pi. Cosine of pi is -1, so 5 minus -1 gives us 5 + 1, which is 6. Now we can rewrite the integral in terms of UNDU. We're integrating from 4 up to 6. Sin TDT becomes DU. And in the denominator, we have 5 minus cosinet which is equal to u. This is now an elementary integral. Its value is ln of the absolute value of u, but because u is between 4 and 6, we can omit the absolute value and simply evaluate ln of u from 4 up to 6. Applying the fundamental theorem of calculus part 2, we get ln of 6 minus LN of 4. And now we can simplify using the properties of logarithms that's ln of 64, which is equal to ln of 3 halves. Which corresponds to the answer of choice b. Thank you for watching.

Evaluate the integrals in Exercises 39–56.43. ∫(from 0 to π)(sin t)/(2 - cos t) dt

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Evaluate the integrals in Exercises 39–56.
43. ∫(from 0 to π)(sin t)/(2 - cos t) dt