If `int(dx)/(p^(2)sin^(2)x+r^(2)cos^(2)x)=(1)/(12)tan^(-1)(3tanx)+c` then the value of p `sinx+rcosx` can be

Options

A. `(6)/(sqrt5)`
B. `sqrt(5)`
C. `6sqrt3`
D. `-4`

Answer

To solve the given problem, we need to evaluate the integral and compare it with the provided expression. Let's go through the solution step by step.

### Step 1: Set up the integral
We start with the integral:
\[ I = \int \frac{dx}{p^2 \sin^2 x + r^2 \cos^2 x} \]

### Step 2: Simplify the integral
To simplify the integral, we divide the numerator and denominator by \(\cos^2 x\):
\[ I = \int \frac{1}{p^2 \tan^2 x + r^2} \sec^2 x \, dx \]

### Step 3: Use substitution
Let \(t = \tan x\). Then, we have:
\[ dx = \frac{dt}{\sec^2 x} = dt \]
Thus, the integral becomes:
\[ I = \int \frac{dt}{p^2 t^2 + r^2} \]

### Step 4: Apply the integral formula
We can use the formula for the integral:
\[ \int \frac{dt}{a^2 + t^2} = \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) + C \]
In our case, \(a^2 = r^2\) and \(t^2 = p^2 t^2\), so we rewrite the integral:
\[ I = \frac{1}{r} \tan^{-1} \left( \frac{t}{\frac{r}{p}} \right) + C \]

### Step 5: Substitute back for \(t\)
Substituting back \(t = \tan x\):
\[ I = \frac{1}{r} \tan^{-1} \left( \frac{p \tan x}{r} \right) + C \]

### Step 6: Compare with the given expression
According to the problem, we have:
\[ I = \frac{1}{12} \tan^{-1}(3 \tan x) + C \]
Now, we compare the two expressions:
\[ \frac{1}{r} \tan^{-1} \left( \frac{p \tan x}{r} \right) = \frac{1}{12} \tan^{-1}(3 \tan x) \]

### Step 7: Equate coefficients
From the comparison, we can equate:
1. \(pr = 12\)
2. \(\frac{p}{r} = 3 \Rightarrow p = 3r\)

### Step 8: Solve for \(p\) and \(r\)
Substituting \(p = 3r\) into \(pr = 12\):
\[ (3r)r = 12 \Rightarrow 3r^2 = 12 \Rightarrow r^2 = 4 \Rightarrow r = \pm 2 \]
Then, substituting back to find \(p\):
\[ p = 3r = 3(\pm 2) = \pm 6 \]

### Step 9: Find \(p \sin x + r \cos x\)
Now we need to find the range of \(p \sin x + r \cos x\):
\[ p \sin x + r \cos x = 6 \sin x + 2 \cos x \]

### Step 10: Determine the range
The expression \(p \sin x + r \cos x\) can be expressed in the form \(R \sin(x + \phi)\) where:
\[ R = \sqrt{p^2 + r^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \]
Thus, the range of \(p \sin x + r \cos x\) is:
\[ [-2\sqrt{10}, 2\sqrt{10}] \]

### Conclusion
The value of \(p \sin x + r \cos x\) can be any value within the range \([-2\sqrt{10}, 2\sqrt{10}]\).