/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 25

Evaluate the following integrals. $$\int e^{-x} \sin 4 x d x$$

Short Answer

Expert verified
Question: Evaluate the integral, $$\int e^{-x} \sin 4x \, dx$$. Answer: The integral can be found by applying integration by parts twice, resulting in the solution $$\int e^{-x}\sin4x \, dx = \frac{1}{17}\left(-\frac{1}{4}e^{-x}\cos4x + \frac{1}{16}e^{-x}\sin4x\right) + C$$.

Step by step solution

01

Choose u and dv

Choose \(u = e^{-x}\), and \(dv = \sin 4x \, dx\). Now, we need to differentiate \(u\) and integrate \(dv\) to find \(du\) and \(v\).
02

Differentiate u and Integrate dv

Differentiate \(u\) with respect to \(x\) and integrate \(dv\): $$du = \frac{d}{dx} (e^{-x}) \, dx = -e^{-x} \, dx$$ $$v = \int \sin 4x \, dx = -\frac{1}{4} \cos 4x + C$$
03

Use Integration by Parts Formula

Now, using the integration by parts formula, we have: $$\int e^{-x} \sin 4x d x = \left(-\frac{1}{4}e^{-x}\cos4x\right) - \int \left(-\frac{1}{4}e^{-x}\cos4x\right) \, dx$$ Let's notice that the remaining integral is again a product of exponential and trigonometric functions. We'll apply integration by parts once again.
04

Apply Integration by Parts Again

Choose \(\tilde{u} = -\frac{1}{4}e^{-x}\) and \(\tilde{dv} = \cos 4x \, dx\). Differentiate \(\tilde{u}\) and integrate \(\tilde{dv}\): $$\tilde{du} = \frac{1}{4}e^{-x} \, dx$$ $$\tilde{v} = \int \cos 4x \, dx = \frac{1}{4}\sin 4x + C$$ Now, substituting into the integration by parts formula, $$\int -\frac{1}{4}e^{-x}\cos4x \, dx = \left(-\frac{1}{16}e^{-x}\sin4x\right) - \int \left(-\frac{1}{16}e^{-x}\sin4x\right) \, dx$$ Multiply the remaining integral by \(-1\): $$\int \frac{1}{16}e^{-x}\sin4x \, dx = \frac{1}{16}\int e^{-x}\sin4x \, dx$$
05

Combine the Results

Notice that we now have the original integral on the right side. So let's combine the results: $$\int e^{-x} \sin 4x d x = -\frac{1}{4}e^{-x}\cos4x - \left(-\frac{1}{16}e^{-x}\sin4x\right) - \frac{1}{16}\int e^{-x}\sin4x \, dx$$ Now, add the remaining integral on both sides: $$\frac{17}{16}\int e^{-x}\sin4x \, dx = -\frac{1}{4}e^{-x}\cos4x + \frac{1}{16}e^{-x}\sin4x$$ Finally, divide by \(\frac{17}{16}\): $$\int e^{-x}\sin4x \, dx = \frac{1}{17}\left(-\frac{1}{4}e^{-x}\cos4x + \frac{1}{16}e^{-x}\sin4x\right) + C$$

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