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Chapter 4: Problem 103

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{e^{2 x}-e^{-2 x}}{2} d x$$

Short Answer

Expert verified
Question: Find the indefinite integral of the following function: $$\int \frac{e^{2 x}-e^{-2 x}}{2} d x$$ Answer: The indefinite integral of the given function is: $$\int \frac{e^{2 x}-e^{-2 x}}{2} d x = \frac{1}{4} e^{2x} - \frac{1}{4} e^{-2x} + C$$

Step by step solution

01

Identify the given function

The given function to integrate is: $$\int \frac{e^{2 x}-e^{-2 x}}{2} d x$$
02

Rewrite the function and integrate separately

We can split the function into two separate integrals and multiply the constants outside the integral: $$\int \frac{e^{2 x}-e^{-2 x}}{2} d x = \frac{1}{2}\int e^{2x} dx - \frac{1}{2}\int e^{-2x} dx$$
03

Integrate each part

Now, we will integrate each part separately. Using substitution, let's solve the first integral by letting \(u = 2x\), then \(du = 2 dx\) and \(\frac{1}{2} du = dx\): $$\frac{1}{2}\int e^{2x} dx = \frac{1}{2}\int e^u \frac{1}{2} du = \frac{1}{4} \int e^u du = \frac{1}{4} e^u + C_1 = \frac{1}{4} e^{2x} + C_1$$ Now, let's solve the second integral by letting \(v = -2x\), then \(dv = -2 dx\) and \(-\frac{1}{2} dv = dx\): $$\frac{1}{2}\int e^{-2x} dx = \frac{1}{2}\int e^v (-\frac{1}{2}) dv = -\frac{1}{4} \int e^v dv = -\frac{1}{4} e^v + C_2 = -\frac{1}{4} e^{-2x} + C_2$$
04

Combine the integrals

Now we combine the two integrals: $$\frac{1}{2}\int e^{2x} dx - \frac{1}{2}\int e^{-2x} dx = \frac{1}{4} e^{2x} + C_1 -\frac{1}{4} e^{-2x} + C_2$$ Combine \(C_1\) and \(C_2\) into a new constant \(C\): $$= \frac{1}{4} e^{2x} - \frac{1}{4} e^{-2x} + C$$
05

Differentiate the result to check the work

Now we will differentiate the result to check if it matches the original function. Using the chain rule, we have: $$\frac{d}{dx}\left(\frac{1}{4} e^{2x} - \frac{1}{4} e^{-2x} + C\right) = \frac{1}{4}(2e^{2x}) - \frac{1}{4}(-2e^{-2x})$$ Simplify: $$= \frac{e^{2 x}-e^{-2 x}}{2}$$ The derivative matches the original function, so our integration is correct. The indefinite integral of the given function is: $$\int \frac{e^{2 x}-e^{-2 x}}{2} d x = \frac{1}{4} e^{2x} - \frac{1}{4} e^{-2x} + C$$

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