Evaluate the definite integral.∫0π3sinθ1+cosθdθ\int_0^{\frac{\pi}{3}}\frac{sin\theta}{1+\cos\theta}d\theta

ln ⁡ ( 4 3 ) ≈ 0.29 \ln\left(\frac43\right)\thickapprox0.29

Evaluate the definite integral. ∫ 0 π 3 s i n θ 1 + cos θ d θ \int_0^{\frac{\pi}{3}}\frac{sin\theta}{1+\cos\theta}d\theta

we're asked to evaluate the definite integral from zero to pi over three of sine of theta over one plus cosine of theta d theta. Now since we're working with a rational function here, we know that a good strategy to use is to do a substitution where u is equal to our denominator. So let's try that here. If u is equal to one plus the cosine of theta, what is du? Well, that's going to be equal to the derivative of that. So negative sine theta d theta. Now we can see that we are on that is almost what we have in our integrand here. We have one plus cosine theta, and then we have sine theta d theta. We're just missing that negative. So I can go ahead and just multiply by negative one in order to make that work. And then if I multiply by negative one, I also need to multiply it by its reciprocal so that I'm not actually changing the value of my integral here. So now I have u and I have du. So I can rewrite this as a negative integral one over u du. Now I'm gonna ignore those bounds for now and stick them on at the end once I have this back in terms of theta. So this will negative one over u du or negative integral one over u du. This is going to give us negative natural log of the absolute value of u plus c. But, again, here, we don't really have to worry about that plus c because we know that this is bounded. So putting this function of theta back in here, this is negative natural log of the absolute value of one plus cosine theta and then bounded from zero to Pi over three. So now that we have this antiderivative here, we want to go ahead and plug in those bounds. So going ahead and doing this here, this is negative. I'm going to keep that negative out front for now. Natural log of one plus cosine Pi over three minus natural log of the absolute value of one plus cosine of zero. So now we wanna go ahead and evaluate those trig functions here. Cosine of PI over three, that's going to give us a one half. So that's really one plus one half and the cosine of zero is equal to one. So I can go ahead and rewrite this based on that. So negative natural log of one plus one half. That is already a positive value, so I don't have to worry about my absolute value signs here. One plus one half is going to be three half. So this is natural log of three halves, and then minus the natural log of one plus one, which is two. Now when we're working with two natural logs being added or subtracted, we can combine them into one natural log based on our log properties. So this is negative natural log of three halves divided by two, which is the same thing as multiplying by one half here. So three halves times one half is three fourths. So this is negative natural log of three fourths, but we can do one additional thing to simplify this here. This negative, I can go ahead and make the exponent of my argument here. So this is really the natural log of three fourths to the power of negative one, which means that I can just flip that fraction into its reciprocal. So this is the natural log of four thirds. So this is an appropriate final answer. But if you are using a calculator and you want your final answer in terms of decimals, you can type this right into your calculator and this will end up being here about 0.29 having rounded that off to two decimal places. Now whenever we're working with natural logs, your answer might look slightly different than mine if you use different log properties to do simplification. But your answer still could be equivalent. So it's important to really understand your log properties so that you can see if you got a correct or incorrect answer. Now you can always ask us in the comments if you have any questions. I'll see you in the next one.

Evaluate the definite integral. ∫ 0 π 3 s i n θ 1 + cos ⁡ θ d θ \int_0^{\frac{\pi}{3}}\frac{sin\theta}{1+\cos\theta}d\theta

ln ⁡ ( 4 3 ) ≈ 0.29 \ln\left(\frac43\right)\thickapprox0.29

ln ⁡ ( 3 4 ) ≈ − 0.29 \ln\left(\frac34\right)\thickapprox-0.29

ln ⁡ ( 2 ) − cos ⁡ ( π 3 ) ln ⁡ ( 1 + cos ⁡ π 3 ) ≈ 1.99 × 10 − 4 \ln\left(2\right)-\cos\left(\frac{\pi}{3}\right)\ln\left(1+\cos\frac{\pi}{3}\right)\thickapprox1.99\times10^{-4}

ln ⁡ ( 2 ) − cos ⁡ ( π 3 ) ln ⁡ ( 1 + cos ⁡ π 3 ) ≈ 0.49 \ln\left(2\right)-\cos\left(\frac{\pi}{3}\right)\ln\left(1+\cos\frac{\pi}{3}\right)\thickapprox0.49

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

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Multiple Choice

Evaluate the definite integral.
$\int_{0}^{\frac{π}{3}} \frac{s i n θ}{1 + \cos θ} d θ$

$l n four thirds almost equals 0.29$

$\ln (\frac{3}{4}) \approx - 0.29$

$\ln (2) - \cos (\frac{π}{3}) \ln (1 + \cos \frac{π}{3}) \approx 1.99 \times 10^{- 4}$

$\ln (2) - \cos (\frac{π}{3}) \ln (1 + \cos \frac{π}{3}) \approx 0.49$

Verified step by step guidance

Step 1: Recognize that the integral is ∫₀^(π/3) (sin(θ) / (1 + cos(θ))) dθ. To simplify this, consider a substitution. Let u = 1 + cos(θ), which implies that du = -sin(θ) dθ.

Step 2: Adjust the limits of integration according to the substitution. When θ = 0, u = 1 + cos(0) = 2. When θ = π/3, u = 1 + cos(π/3) = 1 + 1/2 = 3/2.

Step 3: Rewrite the integral in terms of u. The integral becomes ∫₂^(3/2) (-1/u) du. The negative sign comes from the substitution du = -sin(θ) dθ.

Step 4: Integrate the simplified expression. The integral of -1/u with respect to u is -ln|u|. So, the integral becomes -ln|u| evaluated from u = 2 to u = 3/2.

Step 5: Apply the Fundamental Theorem of Calculus to evaluate the definite integral. Substitute the limits into -ln|u| to get -ln(3/2) + ln(2). Simplify this expression to find the final result.

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