Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
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 appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
