Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Sure, let's solve the integral [tex]\(\int e^{2x} \cos 3x \, dx\)[/tex].
To solve this integral, we will use integration by parts. Integration by parts is given by the formula:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
Here, we'll let:
[tex]\[ u = e^{2x} \quad \text{and} \quad dv = \cos 3x \, dx \][/tex]
Now, we need to find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ du = \frac{d}{dx}(e^{2x}) \, dx = 2e^{2x} \, dx \][/tex]
[tex]\[ v = \int \cos 3x \, dx = \frac{1}{3} \sin 3x \][/tex]
Now, we apply the integration by parts formula:
[tex]\[ \int e^{2x} \cos 3x \, dx = e^{2x} \cdot \frac{1}{3} \sin 3x - \int \left( \frac{1}{3} \sin 3x \right) \cdot 2e^{2x} \, dx \][/tex]
Simplify the expression:
[tex]\[ = \frac{1}{3} e^{2x} \sin 3x - \frac{2}{3} \int e^{2x} \sin 3x \, dx \][/tex]
Now we need to solve the integral [tex]\(\int e^{2x} \sin 3x \, dx\)[/tex]. We can use integration by parts again for this integral. Let:
[tex]\[ u = e^{2x} \quad \text{and} \quad dv = \sin 3x \, dx \][/tex]
Again, find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ du = \frac{d}{dx}(e^{2x}) \, dx = 2e^{2x} \, dx \][/tex]
[tex]\[ v = \int \sin 3x \, dx = -\frac{1}{3} \cos 3x \][/tex]
Applying integration by parts formula:
[tex]\[ \int e^{2x} \sin 3x \, dx = e^{2x} \cdot \left(-\frac{1}{3} \cos 3x\right) - \int \left(-\frac{1}{3} \cos 3x \right) \cdot 2e^{2x} \, dx \][/tex]
Simplify the expression:
[tex]\[ = -\frac{1}{3} e^{2x} \cos 3x + \frac{2}{3} \int e^{2x} \cos 3x \, dx \][/tex]
Now let's denote [tex]\(\int e^{2x} \cos 3x \, dx\)[/tex] as [tex]\(I\)[/tex]. Then from the above steps:
[tex]\[ I = \frac{1}{3} e^{2x} \sin 3x - \frac{2}{3} \left( -\frac{1}{3} e^{2x} \cos 3x + \frac{2}{3} I \right) \][/tex]
[tex]\[ I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x - \frac{4}{9} I \][/tex]
Combine the [tex]\(I\)[/tex] terms on one side:
[tex]\[ I + \frac{4}{9} I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
[tex]\[ \left( 1 + \frac{4}{9} \right) I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
[tex]\[ \frac{13}{9} I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
Solve for [tex]\(I\)[/tex]:
[tex]\[ I = \frac{9}{13} \left( \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \right) \][/tex]
[tex]\[ I = \frac{3}{13} e^{2x} \sin 3x + \frac{2}{13} e^{2x} \cos 3x \][/tex]
Hence, the integral [tex]\(\int e^{2x} \cos 3x \, dx\)[/tex] is:
[tex]\[ \int e^{2x} \cos 3x \, dx = \frac{3}{13} e^{2x} \sin 3x + \frac{2}{13} e^{2x} \cos 3x + C \][/tex]
Where [tex]\(C\)[/tex] is the constant of integration.
To solve this integral, we will use integration by parts. Integration by parts is given by the formula:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
Here, we'll let:
[tex]\[ u = e^{2x} \quad \text{and} \quad dv = \cos 3x \, dx \][/tex]
Now, we need to find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ du = \frac{d}{dx}(e^{2x}) \, dx = 2e^{2x} \, dx \][/tex]
[tex]\[ v = \int \cos 3x \, dx = \frac{1}{3} \sin 3x \][/tex]
Now, we apply the integration by parts formula:
[tex]\[ \int e^{2x} \cos 3x \, dx = e^{2x} \cdot \frac{1}{3} \sin 3x - \int \left( \frac{1}{3} \sin 3x \right) \cdot 2e^{2x} \, dx \][/tex]
Simplify the expression:
[tex]\[ = \frac{1}{3} e^{2x} \sin 3x - \frac{2}{3} \int e^{2x} \sin 3x \, dx \][/tex]
Now we need to solve the integral [tex]\(\int e^{2x} \sin 3x \, dx\)[/tex]. We can use integration by parts again for this integral. Let:
[tex]\[ u = e^{2x} \quad \text{and} \quad dv = \sin 3x \, dx \][/tex]
Again, find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ du = \frac{d}{dx}(e^{2x}) \, dx = 2e^{2x} \, dx \][/tex]
[tex]\[ v = \int \sin 3x \, dx = -\frac{1}{3} \cos 3x \][/tex]
Applying integration by parts formula:
[tex]\[ \int e^{2x} \sin 3x \, dx = e^{2x} \cdot \left(-\frac{1}{3} \cos 3x\right) - \int \left(-\frac{1}{3} \cos 3x \right) \cdot 2e^{2x} \, dx \][/tex]
Simplify the expression:
[tex]\[ = -\frac{1}{3} e^{2x} \cos 3x + \frac{2}{3} \int e^{2x} \cos 3x \, dx \][/tex]
Now let's denote [tex]\(\int e^{2x} \cos 3x \, dx\)[/tex] as [tex]\(I\)[/tex]. Then from the above steps:
[tex]\[ I = \frac{1}{3} e^{2x} \sin 3x - \frac{2}{3} \left( -\frac{1}{3} e^{2x} \cos 3x + \frac{2}{3} I \right) \][/tex]
[tex]\[ I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x - \frac{4}{9} I \][/tex]
Combine the [tex]\(I\)[/tex] terms on one side:
[tex]\[ I + \frac{4}{9} I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
[tex]\[ \left( 1 + \frac{4}{9} \right) I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
[tex]\[ \frac{13}{9} I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
Solve for [tex]\(I\)[/tex]:
[tex]\[ I = \frac{9}{13} \left( \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \right) \][/tex]
[tex]\[ I = \frac{3}{13} e^{2x} \sin 3x + \frac{2}{13} e^{2x} \cos 3x \][/tex]
Hence, the integral [tex]\(\int e^{2x} \cos 3x \, dx\)[/tex] is:
[tex]\[ \int e^{2x} \cos 3x \, dx = \frac{3}{13} e^{2x} \sin 3x + \frac{2}{13} e^{2x} \cos 3x + C \][/tex]
Where [tex]\(C\)[/tex] is the constant of integration.
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
