Answer :
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.
