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Chapter 4: Problem 104

Determine the following indefinite integrals. Check your work by differentiation. $$\int(4 \cos 4 w-3 \sin 3 w) d w$$

Short Answer

Expert verified
Question: Find the indefinite integral of the function \(4\cos 4w - 3\sin 3w\) with respect to \(w\). Answer: The indefinite integral of the given function is \(\int(4\cos 4w - 3\sin 3w) dw = \sin 4w - \cos 3w + C\), where \(C\) is the constant of integration.

Step by step solution

01

Identify parts to integrate individually

The given integral is a linear combination of two functions, so we can integrate each part separately: $$\int (4\cos 4w - 3\sin 3w) dw = 4\int\cos 4w \, dw - 3\int\sin 3w \, dw.$$
02

Integrate first function

To integrate the first function, use the formula: $$\int \cos ax \, dx = \frac{1}{a}\sin ax + C_1.$$ Here, \(a = 4\). So, $$4\int\cos 4w \, dw = 4\cdot\frac{1}{4}\sin 4w + C_1 = \sin 4w + C_1$$
03

Integrate second function

To integrate the second function, use the formula: $$\int\sin ax \, dx = -\frac{1}{a}\cos ax + C_2.$$ Here, \(a = 3\). So, $$3\int\sin 3w \, dw = -3\cdot\frac{1}{3}\cos 3w + C_2 = -\cos 3w + C_2$$
04

Combine the results of both integrations

Now that we have integrated both parts, we can combine the results: $$\int(4\cos 4w - 3\sin 3w) dw = (\sin 4w + C_1) - (\cos 3w + C_2) = \sin 4w - \cos 3w + C,$$ where \(C = C_1 - C_2\).
05

Differentiate the result to check our work

Now we differentiate the integral to ensure we obtain the original function back: $$\frac{d}{dw}(\sin 4w - \cos 3w + C) = 4\cos 4w + 3\sin 3w.$$ Since the result of the differentiation is the original function, this confirms our integration is correct. Therefore, the indefinite integral is: $$\int(4\cos 4w - 3\sin 3w) dw = \sin 4w - \cos 3w + C.$$

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