If `u=inte^(ax)sin " bx dx" and v=int(e^(ax))cos " bx dx"`,then `tan^(-1)((u)/(v))+tan^(-1)((b)/(a))` equals

Options

A. bx
B. 2 bx
C. `b^(2)x^(2)`
D. `sqrt(bx)`

Explanation

To solve the problem, we need to find the value of \( \tan^{-1}\left(\frac{u}{v}\right) + \tan^{-1}\left(\frac{b}{a}\right) \) given:

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

### Step 1: Use the Direct Formulas for \( u \) and \( v \)

The direct formulas for the integrals are:

\[ u = \int e^{ax} \sin(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) \]

\[ v = \int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) \]

### Step 2: Substitute \( u \) and \( v \) into \( \tan^{-1}\left(\frac{u}{v}\right) \)

Now, we substitute \( u \) and \( v \) into the expression \( \tan^{-1}\left(\frac{u}{v}\right) \):

\[ \frac{u}{v} = \frac{\frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx))}{\frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx))} \]

The \( \frac{e^{ax}}{a^2 + b^2} \) terms cancel out:

\[ \frac{u}{v} = \frac{a \sin(bx) - b \cos(bx)}{a \cos(bx) + b \sin(bx)} \]

### Step 3: Use the Tangent Addition Formula

Recognizing that the expression can be rewritten using the tangent addition formula, we can express it as:

\[ \tan^{-1}\left(\frac{u}{v}\right) = \tan^{-1}\left(\frac{a \sin(bx) - b \cos(bx)}{a \cos(bx) + b \sin(bx)}\right) \]

This can be interpreted as \( \tan^{-1}(\tan(\theta)) \) where \( \theta = bx - \phi \) for some angle \( \phi \).

### Step 4: Find \( \tan^{-1}\left(\frac{b}{a}\right) \)

Next, we calculate \( \tan^{-1}\left(\frac{b}{a}\right) \). This is simply:

\[ \tan^{-1}\left(\frac{b}{a}\right) = \phi \]

### Step 5: Combine the Results

Now, we combine the two results:

\[ \tan^{-1}\left(\frac{u}{v}\right) + \tan^{-1}\left(\frac{b}{a}\right) = (bx - \phi) + \phi = bx \]

### Final Result

Thus, we conclude that:

\[ \tan^{-1}\left(\frac{u}{v}\right) + \tan^{-1}\left(\frac{b}{a}\right) = bx \]