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Chapter 17: Problem 11

Find \(\int \frac{\sin x+\cos x}{2} \mathrm{~d} x\).

Short Answer

Expert verified
Answer: The result of the integral is \(-\frac{1}{2} \cos x + \frac{1}{2} \sin x + C\), where \(C\) is the constant of integration.

Step by step solution

01

Simplify the function

We can rewrite the given integral as a sum of two simpler integrals: $$\int \frac{\sin x+\cos x}{2} \mathrm{~d} x = \frac{1}{2} \int \sin x \mathrm{~d} x + \frac{1}{2} \int \cos x \mathrm{~d} x.$$
02

Integrate each term separately

Now, we will apply the integral rules to each term: $$\frac{1}{2} \int \sin x \mathrm{~d} x = -\frac{1}{2} \cos x + C_1,$$ $$\frac{1}{2} \int \cos x \mathrm{~d} x = \frac{1}{2} \sin x + C_2,$$ where \(C_1\) and \(C_2\) are the constants of integration.
03

Combine the results

Combine the results from Step 2, we get $$\int \frac{\sin x+\cos x}{2} \mathrm{~d} x = -\frac{1}{2} \cos x + \frac{1}{2} \sin x + C_1 + C_2.$$
04

Add the constant of integration

Since \(C_1\) and \(C_2\) are both constants of integration, we can combine them into a single constant, denoted as \(C\): $$\int \frac{\sin x+\cos x}{2} \mathrm{~d} x = -\frac{1}{2} \cos x + \frac{1}{2} \sin x + C.$$ Hence, the solution to the given integral is $$\int \frac{\sin x+\cos x}{2} \mathrm{~d} x = -\frac{1}{2} \cos x + \frac{1}{2} \sin x + C.$$

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