Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
To solve the integral [tex]\(\int e^x \cos(6x) \, dx\)[/tex], we can use the method of integration by parts. Integration by parts is a technique based on the product rule of differentiation and is given by:
[tex]\[ \int u \, dv = uv - \int v \, du. \][/tex]
For this problem, we will need to apply integration by parts twice. Let's set it up as follows:
1. First application of integration by parts:
Let [tex]\(u = e^x \)[/tex] and [tex]\( dv = \cos(6x) \, dx \)[/tex].
Then,
[tex]\[ du = e^x \, dx \quad \text{and} \quad v = \int \cos(6x) \, dx. \][/tex]
To find [tex]\(v\)[/tex], we integrate [tex]\(\cos(6x)\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ v = \int \cos(6x) \, dx = \frac{1}{6} \sin(6x). \][/tex]
Now apply integration by parts:
[tex]\[ \int e^x \cos(6x) \, dx = e^x \cdot \frac{1}{6} \sin(6x) - \int \frac{1}{6} \sin(6x) \cdot e^x \, dx. \][/tex]
Simplify to:
[tex]\[ \int e^x \cos(6x) \, dx = \frac{1}{6} e^x \sin(6x) - \frac{1}{6} \int e^x \sin(6x) \, dx. \][/tex]
2. Second application of integration by parts:
Now, we need to evaluate [tex]\(\int e^x \sin(6x) \, dx \)[/tex]. We again use integration by parts for this integral.
Let [tex]\(u = e^x \)[/tex] and [tex]\( dv = \sin(6x) \, dx \)[/tex].
Then,
[tex]\[ du = e^x \, dx \quad \text{and} \quad v = \int \sin(6x) \, dx. \][/tex]
To find [tex]\(v\)[/tex], we integrate [tex]\(\sin(6x)\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ v = \int \sin(6x) \, dx = -\frac{1}{6} \cos(6x). \][/tex]
Now apply integration by parts again:
[tex]\[ \int e^x \sin(6x) \, dx = e^x \cdot \left( -\frac{1}{6} \cos(6x) \right) - \int \left( -\frac{1}{6} \cos(6x) \right) e^x \, dx. \][/tex]
Simplify to:
[tex]\[ \int e^x \sin(6x) \, dx = -\frac{1}{6} e^x \cos(6x) + \frac{1}{6} \int e^x \cos(6x) \, dx. \][/tex]
Let's denote [tex]\(\int e^x \cos(6x) \, dx\)[/tex] as [tex]\(I\)[/tex]:
[tex]\[ I = \frac{1}{6} e^x \sin(6x) - \frac{1}{6} \left( -\frac{1}{6} e^x \cos(6x) + \frac{1}{6} I \right). \][/tex]
Simplify further:
[tex]\[ I = \frac{1}{6} e^x \sin(6x) + \frac{1}{36} e^x \cos(6x) - \frac{1}{36} I. \][/tex]
Multiply through by 36 to clear the fraction:
[tex]\[ 36I = 6e^x \sin(6x) + e^x \cos(6x) - I. \][/tex]
Add [tex]\(I\)[/tex] to both sides:
[tex]\[ 37I = 6e^x \sin(6x) + e^x \cos(6x). \][/tex]
Finally, solve for [tex]\(I\)[/tex]:
[tex]\[ I = \int e^x \cos(6x) \, dx = \frac{6e^x \sin(6x) + e^x \cos(6x)}{37}. \][/tex]
Thus, the solution to the integral is:
[tex]\[ \int e^x \cos(6x) \, dx = \frac{6 e^x \sin(6x) + e^x \cos(6x)}{37} + C, \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
[tex]\[ \int u \, dv = uv - \int v \, du. \][/tex]
For this problem, we will need to apply integration by parts twice. Let's set it up as follows:
1. First application of integration by parts:
Let [tex]\(u = e^x \)[/tex] and [tex]\( dv = \cos(6x) \, dx \)[/tex].
Then,
[tex]\[ du = e^x \, dx \quad \text{and} \quad v = \int \cos(6x) \, dx. \][/tex]
To find [tex]\(v\)[/tex], we integrate [tex]\(\cos(6x)\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ v = \int \cos(6x) \, dx = \frac{1}{6} \sin(6x). \][/tex]
Now apply integration by parts:
[tex]\[ \int e^x \cos(6x) \, dx = e^x \cdot \frac{1}{6} \sin(6x) - \int \frac{1}{6} \sin(6x) \cdot e^x \, dx. \][/tex]
Simplify to:
[tex]\[ \int e^x \cos(6x) \, dx = \frac{1}{6} e^x \sin(6x) - \frac{1}{6} \int e^x \sin(6x) \, dx. \][/tex]
2. Second application of integration by parts:
Now, we need to evaluate [tex]\(\int e^x \sin(6x) \, dx \)[/tex]. We again use integration by parts for this integral.
Let [tex]\(u = e^x \)[/tex] and [tex]\( dv = \sin(6x) \, dx \)[/tex].
Then,
[tex]\[ du = e^x \, dx \quad \text{and} \quad v = \int \sin(6x) \, dx. \][/tex]
To find [tex]\(v\)[/tex], we integrate [tex]\(\sin(6x)\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ v = \int \sin(6x) \, dx = -\frac{1}{6} \cos(6x). \][/tex]
Now apply integration by parts again:
[tex]\[ \int e^x \sin(6x) \, dx = e^x \cdot \left( -\frac{1}{6} \cos(6x) \right) - \int \left( -\frac{1}{6} \cos(6x) \right) e^x \, dx. \][/tex]
Simplify to:
[tex]\[ \int e^x \sin(6x) \, dx = -\frac{1}{6} e^x \cos(6x) + \frac{1}{6} \int e^x \cos(6x) \, dx. \][/tex]
Let's denote [tex]\(\int e^x \cos(6x) \, dx\)[/tex] as [tex]\(I\)[/tex]:
[tex]\[ I = \frac{1}{6} e^x \sin(6x) - \frac{1}{6} \left( -\frac{1}{6} e^x \cos(6x) + \frac{1}{6} I \right). \][/tex]
Simplify further:
[tex]\[ I = \frac{1}{6} e^x \sin(6x) + \frac{1}{36} e^x \cos(6x) - \frac{1}{36} I. \][/tex]
Multiply through by 36 to clear the fraction:
[tex]\[ 36I = 6e^x \sin(6x) + e^x \cos(6x) - I. \][/tex]
Add [tex]\(I\)[/tex] to both sides:
[tex]\[ 37I = 6e^x \sin(6x) + e^x \cos(6x). \][/tex]
Finally, solve for [tex]\(I\)[/tex]:
[tex]\[ I = \int e^x \cos(6x) \, dx = \frac{6e^x \sin(6x) + e^x \cos(6x)}{37}. \][/tex]
Thus, the solution to the integral is:
[tex]\[ \int e^x \cos(6x) \, dx = \frac{6 e^x \sin(6x) + e^x \cos(6x)}{37} + C, \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
