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Chapter 4: Problem 38

Determine the following indefinite integrals. Check your work by differentiation. $$\int\left(\sin 4 t-\sin \frac{t}{4}\right) d t$$

Short Answer

Expert verified
Answer: The indefinite integral of the function \(\sin(4t) - \sin(\frac{t}{4})\) is \(-\frac{1}{4}\cos(4t) + 4\cos\left(\frac{t}{4}\right) + C\), where \(C\) is the constant of integration.

Step by step solution

01

Break down the integral

Break down the integral into two separate integrals: $$\int\left(\sin 4 t-\sin \frac{t}{4}\right) d t = \int \sin 4 t\, dt - \int \sin \frac{t}{4}\, dt$$
02

Integrate the first function

Integrate \(\sin 4t\) using the formula \(\int \sin(ax+b) dx = -\frac{1}{a}\cos(ax+b) + C\): $$\int \sin 4t\, dt = -\frac{1}{4}\cos(4t) + C_1$$
03

Integrate the second function

Integrate \(\sin \frac{t}{4}\) using the same formula: $$\int \sin \frac{t}{4}\, dt = -\frac{1}{\frac{1}{4}} \cos\left(\frac{t}{4}\right) + C_2 = -4\cos\left(\frac{t}{4}\right) + C_2$$
04

Combine the integration results

Now, combine the results from Step 2 and Step 3: $$\int\left(\sin 4 t-\sin \frac{t}{4}\right) d t = \left(-\frac{1}{4}\cos(4t) + C_1\right) - \left(-4\cos\left(\frac{t}{4}\right) + C_2\right)$$ Simplify the expression to get the final answer: $$\int\left(\sin 4 t-\sin \frac{t}{4}\right) d t = -\frac{1}{4}\cos(4t) + 4\cos\left(\frac{t}{4}\right) + C$$ where \(C = C_1 - C_2\).
05

Check the result by differentiation

We will now take the derivative of our result and check if it matches the original function: Using the chain rule, differentiate the expression: $$\frac{d}{dt}\left(-\frac{1}{4}\cos(4t) + 4\cos\left(\frac{t}{4}\right) + C\right) = \sin(4t) - \sin\left(\frac{t}{4}\right)$$ As the derivative matches the original function, our result is correct: $$\int\left(\sin 4 t-\sin \frac{t}{4}\right) d t = -\frac{1}{4}\cos(4t) + 4\cos\left(\frac{t}{4}\right) + C$$

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