If `u=inte^(ax)sin " bx dx" and v=inte^(ax)cos " bx dx then "(u^(2)+v^(2))(a^(2)+b^(2))=`

Options

A. `2e^(ax)`
B. `e^(2ax)`
C. `2e^(2ax)`
D. `bxe^(ax)`

Explanation

To solve the problem, we need to find the value of \( (u^2 + v^2)(a^2 + b^2) \) where:

- \( u = \int e^{ax} \sin(bx) \, dx \)
- \( v = \int e^{ax} \cos(bx) \, dx \)

### Step 1: Finding \( u \)

We will use integration by parts to evaluate \( u \).

1. Let \( I = \int e^{ax} \sin(bx) \, dx \).
2. Using integration by parts, we set:
- \( f = \sin(bx) \) and \( dg = e^{ax} \, dx \)
- Then, \( df = b \cos(bx) \, dx \) and \( g = \frac{e^{ax}}{a} \)

Using integration by parts:
\[ I = \frac{e^{ax}}{a} \sin(bx) - \int \frac{e^{ax}}{a} b \cos(bx) \, dx \]
\[ I = \frac{e^{ax}}{a} \sin(bx) - \frac{b}{a} \int e^{ax} \cos(bx) \, dx \]

Let \( J = \int e^{ax} \cos(bx) \, dx \). Thus,
\[ I = \frac{e^{ax}}{a} \sin(bx) - \frac{b}{a} J \]

### Step 2: Finding \( J \)

Now we will evaluate \( J \) using integration by parts again.

1. Let \( J = \int e^{ax} \cos(bx) \, dx \).
2. Using integration by parts:
- \( f = \cos(bx) \) and \( dg = e^{ax} \, dx \)
- Then, \( df = -b \sin(bx) \, dx \) and \( g = \frac{e^{ax}}{a} \)

Using integration by parts:
\[ J = \frac{e^{ax}}{a} \cos(bx) + \frac{b}{a} \int e^{ax} \sin(bx) \, dx \]
\[ J = \frac{e^{ax}}{a} \cos(bx) + \frac{b}{a} I \]

### Step 3: Solving the system of equations

Now we have two equations:
1. \( I = \frac{e^{ax}}{a} \sin(bx) - \frac{b}{a} J \)
2. \( J = \frac{e^{ax}}{a} \cos(bx) + \frac{b}{a} I \)

Substituting \( J \) from the second equation into the first:
\[ I = \frac{e^{ax}}{a} \sin(bx) - \frac{b}{a} \left( \frac{e^{ax}}{a} \cos(bx) + \frac{b}{a} I \right) \]
\[ I + \frac{b^2}{a^2} I = \frac{e^{ax}}{a} \sin(bx) - \frac{b}{a^2} e^{ax} \cos(bx) \]
\[ I \left( 1 + \frac{b^2}{a^2} \right) = \frac{e^{ax}}{a} \sin(bx) - \frac{b}{a^2} e^{ax} \cos(bx) \]
\[ I = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) \]

Thus, we have:
\[ u = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) \]

### Step 4: Finding \( v \)

Using similar steps for \( v \):
\[ v = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) \]

### Step 5: Finding \( u^2 + v^2 \)

Now we need to find \( u^2 + v^2 \):
\[ u^2 + v^2 = \left( \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) \right)^2 + \left( \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) \right)^2 \]
Factoring out \( \left( \frac{e^{ax}}{a^2 + b^2} \right)^2 \):
\[ u^2 + v^2 = \frac{e^{2ax}}{(a^2 + b^2)^2} \left( (a \sin(bx) - b \cos(bx))^2 + (a \cos(bx) + b \sin(bx))^2 \right) \]

### Step 6: Simplifying the expression

Expanding the terms inside the parentheses:
\[ = \frac{e^{2ax}}{(a^2 + b^2)^2} \left( a^2 \sin^2(bx) - 2ab \sin(bx) \cos(bx) + b^2 \cos^2(bx) + a^2 \cos^2(bx) + 2ab \sin(bx) \cos(bx) + b^2 \sin^2(bx) \right) \]
The cross terms cancel out:
\[ = \frac{e^{2ax}}{(a^2 + b^2)^2} \left( a^2 (\sin^2(bx) + \cos^2(bx)) + b^2 (\sin^2(bx) + \cos^2(bx)) \right) \]
Using \( \sin^2(bx) + \cos^2(bx) = 1 \):
\[ = \frac{e^{2ax}}{(a^2 + b^2)^2} (a^2 + b^2) \]

### Step 7: Final Result

Now we compute \( (u^2 + v^2)(a^2 + b^2) \):
\[ (u^2 + v^2)(a^2 + b^2) = \frac{e^{2ax}}{(a^2 + b^2)^2} (a^2 + b^2) (a^2 + b^2) = e^{2ax} \cdot 1 = 2 e^{2ax} \]

Thus, the final answer is:
\[ \boxed{2 e^{2ax}} \]