Evaluate the integrals in Exercises 1–22.
∫₀^π sin⁵(x/2)...

Question

Evaluate the integrals in Exercises 1–22.

∫₀^π sin⁵(x/2) dx

Accepted Answer

Welcome back everyone. Find the value of the integral from 0 to pi of sine cubed x dx. A 3 halves, B 4/3, C 2/3, and D8/3. So, for this problem, what we're going to do is first of we'll analyze the function that we're integrating and that's sin cube x. What we can do is simply split the exponent into two parts using the properties of exponents. We can rewrite it as sin squared x multiplied by sin x, and from here we're going to apply the Pythagorean identity. We, we know that sin squared x is a 1 minus cosine squared x. And we're multiplying by sin x. Now, let's rewrite the integral in terms of this form. We get integral from 0 to pi of 1 minus cosine square x multiplied by sin xDX. And what we can do is someplace that. A substitution u equals cosine x. Remember that the next step is to find the derivative of U with respect to X, which is the derivative of cosine and that's negative sine x. And now if we multiply both sides by negative DX, we get negative DU is equal to sin xDX. The reason why we choose this multiplication is because we have sin x dx within the integral, right? So now we can almost rewrite the integral in terms of UNDU but we also want to change the limits of integration. Starting with the lower limit of integration, EU1 is going to be cosine of 0. And now cosine of 0 can be rewritten as divided by 2. And then we are going to write our. Next limit of integration, I'm sorry, cosine of 0 is going to be equal to 1. I'm sorry. And now for the upper limit of integration we have u2 equals cosine of pi, which is -1. So the lower limit of integration is going to be cosine of 0, which is 1, and the upper limit of integration is cosine of pi, which is -1. So now we can rewrite the integral as the integral from 1 up to -1. Of 1 minus cosine 2, that's 1 minus u 2. Multiplied by sin x dx, which is negative duu, so we can just write DU and we can factor out the negative sine. And then we can reverse the limits of integration, that would be the integral from -1 up to 1 and remove the negative sign. This is one of the properties of integrals and simply integrate 1 minus u. Squared d We can apply the power rule. The integral of 1 duu is the integral of DU, which is U, and the integral of u squad is u cubed divided by 3. And we want to evaluate the result from -1 up to 1. That's going to be 1 minus 1 cube divided by 3, that's. 1/3 And we are subtracting the value at -1. So -1 minus -1 cubed gives us -1. So -1 gives us 1 divided by 3. Now let's compute the result 1 minus -1, that's 1 + 1, and we get 2. 1/3 minus 1/3 gives us 2/3, and we can now find the common denominator which is 3, right? So 2 multiplied by 3 gives us 6, and we're subtracting 2. This is going to be divided by 3. 6 minus 2 is 4, so we get 4/3, which corresponds to the answer of choice b. Thank you for watching.

Evaluate the integrals in Exercises 1–22.∫₀^π sin⁵(x/2) dx

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Evaluate the integrals in Exercises 1–22.
∫₀^π sin⁵(x/2) dx