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Chapter 2: Problem 6

Given \(\int_{0}^{1} \frac{\sin t}{1+t} \mathrm{dt}=\alpha\), find \(\int_{4 \pi-2}^{4 \pi} \frac{\sin \frac{\mathrm{t}}{2}}{4 \pi+2-\mathrm{t}} \mathrm{dt}\) in terms of \(\alpha\).

Short Answer

Expert verified
Answer: The value of the integral is \(\int_{4\pi-2}^{4\pi} {\frac{\sin(\frac{t}{2})}{4\pi + 2 - t} \mathrm{dt}} = 2\alpha\).

Step by step solution

01

Find an appropriate substitution

We want to replace \(t\) with a new variable \(u\) in such a way that \(\sin(t)\) becomes \(\sin(\frac{t}{2})\). A natural choice for this substitution is \(u = \frac{t}{2}\). Then, \(\sin(u) = \sin(\frac{t}{2})\). Now, let's find the relationship between the two limits of integration.
02

Analyze the limits of integration

For our substitution \(u = \frac{t}{2}\), the limits of integration change as follows: 1. Lower limit: when \(t = 4\pi - 2\), we have \(u = \frac{4\pi-2}{2} = 2\pi - 1\). 2. Upper limit: when \(t = 4\pi\), we have \(u = \frac{4\pi}{2} = 2\pi\). Now we have the new limits of integration in terms of the variable \(u\).
03

Analyze the function inside the integral

We have \(\frac{\sin t}{1+t} = \frac{\sin(2u)}{4 \pi+2-(2u)}\). We need to differentiate \(t\) with respect to \(u\): \(\frac{dt}{du} = 2\). Then, \(dt = 2du\).
04

Perform the substitution

Now, let's substitute our values into the integral: \(\int_{4\pi-2}^{4\pi}{\frac{\sin(\frac{t}{2})}{4\pi + 2 - t} \mathrm{dt}} = \int_{2\pi-1}^{2\pi} {\frac{\sin(u)}{1+2u} (2\mathrm{du})} = 2\int_{2\pi-1}^{2\pi} {\frac{\sin(u)}{1+2u} \mathrm{du}}\).
05

Rearrange the new integral to match the given integral \(\alpha\)

Our new integral looks similar to the given integral, but with its limits of integration shifted rightward by \(2\pi-1\). We can rearrange this integral by adjusting the limits of integration: \(2\int_{2\pi-1}^{2\pi} {\frac{\sin(u)}{1+2u} \mathrm{du}} = 2\int_{0}^{1} {\frac{\sin(u+2\pi-1)}{1+2(u+2\pi-1)} \mathrm{du}}\). Notice that since \(\sin(u+2\pi) = \sin(u)\), we can simplify the integral: \(2\int_{0}^{1} {\frac{\sin(u+2\pi-1)}{1+2(u+2\pi-1)} \mathrm{du}} = 2\int_{0}^{1} {\frac{\sin(u-1)}{1+2(u-1)} \mathrm{du}}\).
06

Relate the new integral to the given integral \(\alpha\)

Now, compare the integral found above to the given integral: \(2\int_{0}^{1} {\frac{\sin(u-1)}{1+2(u-1)} \mathrm{du}} \quad \text{and} \quad \int_{0}^{1} {\frac{\sin t}{1+t} \mathrm{dt}} = \alpha\). Since the denominators in these integrals have the same form, and the numerators are equivalent up to a shift of the argument of the sine function (which doesn't affect the value of the integral), we can conclude that: \(2\int_{0}^{1} {\frac{\sin(u-1)}{1+2(u-1)} \mathrm{du}} = 2\alpha\).
07

Write the final answer

Finally, we have found that: \(\int_{4\pi-2}^{4\pi} {\frac{\sin(\frac{t}{2})}{4\pi + 2 - t} \mathrm{dt}} = 2\alpha\).

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