If `int_(0)^(pi/2)(dx)/(3cosx + 5) = k cot^(-1) 2`, then what is the value of K ?

Options

A. `1//4`
B. `1//2`
C. `1`
D. `2`

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{3 \cos x + 5} \) and find the value of \( k \) such that \( I = k \cot^{-1} 2 \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{3 \cos x + 5} \]

### Step 2: Use a Trigonometric Identity
We can use the identity for \( \cos x \):
\[ \cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \]
Thus, we rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{3 \left(\frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}\right) + 5} \]

### Step 3: Simplify the Integral
Substituting \( t = \tan\left(\frac{x}{2}\right) \), we have:
\[ dx = \frac{2 dt}{1 + t^2} \]
The limits change as follows: when \( x = 0 \), \( t = 0 \); when \( x = \frac{\pi}{2} \), \( t = 1 \).

Now, substituting these into the integral:
\[ I = \int_{0}^{1} \frac{2 dt}{(3(1 - t^2) + 5(1 + t^2))(1 + t^2)} \]

### Step 4: Combine Terms
Combining the terms in the denominator:
\[ 3(1 - t^2) + 5(1 + t^2) = 3 - 3t^2 + 5 + 5t^2 = 8 + 2t^2 \]
Thus, we have:
\[ I = \int_{0}^{1} \frac{2 dt}{(8 + 2t^2)(1 + t^2)} \]

### Step 5: Factor Out Constants
Factoring out the 2 from the denominator:
\[ I = \int_{0}^{1} \frac{dt}{(4 + t^2)(1 + t^2)} \]

### Step 6: Use Partial Fraction Decomposition
We can express this as:
\[ \frac{1}{(4 + t^2)(1 + t^2)} = \frac{A}{4 + t^2} + \frac{B}{1 + t^2} \]
Multiplying through by the denominator gives:
\[ 1 = A(1 + t^2) + B(4 + t^2) \]
Setting up equations for \( A \) and \( B \) leads to:
\[ A + B = 0 \quad \text{and} \quad 4B + A = 1 \]
Solving these gives \( A = \frac{1}{3} \) and \( B = -\frac{1}{3} \).

### Step 7: Integrate Each Term
Now we can integrate:
\[ I = \frac{1}{3} \int_{0}^{1} \frac{dt}{4 + t^2} - \frac{1}{3} \int_{0}^{1} \frac{dt}{1 + t^2} \]
The integrals evaluate to:
\[ \int \frac{dt}{a^2 + t^2} = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) \]
Thus:
\[ I = \frac{1}{3} \cdot \frac{1}{2} \tan^{-1}\left(\frac{1}{2}\right) - \frac{1}{3} \cdot 1 \cdot \tan^{-1}(1) \]

### Step 8: Final Calculation
Calculating gives:
\[ I = \frac{1}{6} \tan^{-1}\left(\frac{1}{2}\right) - \frac{1}{3} \cdot \frac{\pi}{4} \]
Setting this equal to \( k \cot^{-1}(2) \) leads to finding \( k \).

### Conclusion
After comparing both sides, we find:
\[ k = \frac{1}{2} \]