IF `int_0^(pi//2) (dx)/(3 cos x+5) =k cot^-1 2`. Then what is the value of k

Options

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

Answer

To solve the integral \( \int_0^{\frac{\pi}{2}} \frac{dx}{3 \cos x + 5} = k \cot^{-1} 2 \), we will follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_0^{\frac{\pi}{2}} \frac{dx}{3 \cos x + 5} \]

### Step 2: Simplify the denominator
We can rewrite the denominator:
\[ 3 \cos x + 5 = 5 + 3 \cos x \]

### Step 3: Use a trigonometric substitution
We can use the substitution \( t = \tan\left(\frac{x}{2}\right) \). This gives us:
\[ \cos x = \frac{1 - t^2}{1 + t^2} \quad \text{and} \quad dx = \frac{2}{1 + t^2} dt \]
The limits change as follows:
- When \( x = 0 \), \( t = 0 \)
- When \( x = \frac{\pi}{2} \), \( t = 1 \)

### Step 4: Substitute into the integral
Substituting these into the integral, we have:
\[ I = \int_0^1 \frac{2}{1 + t^2} \cdot \frac{1}{3 \cdot \frac{1 - t^2}{1 + t^2} + 5} dt \]

### Step 5: Simplify the integrand
The denominator becomes:
\[ 3 \cdot \frac{1 - t^2}{1 + t^2} + 5 = \frac{3(1 - t^2) + 5(1 + t^2)}{1 + t^2} = \frac{(3 + 5) + (5 - 3)t^2}{1 + t^2} = \frac{8 + 2t^2}{1 + t^2} \]
Thus, the integral simplifies to:
\[ I = \int_0^1 \frac{2(1 + t^2)}{8 + 2t^2} dt = \int_0^1 \frac{2}{2} \cdot \frac{1 + t^2}{4 + t^2} dt = \int_0^1 \frac{1 + t^2}{4 + t^2} dt \]

### Step 6: Split the integral
We can split the integral:
\[ I = \int_0^1 \frac{1}{4 + t^2} dt + \int_0^1 \frac{t^2}{4 + t^2} dt \]

### Step 7: Evaluate the first integral
The first integral can be evaluated as:
\[ \int_0^1 \frac{1}{4 + t^2} dt = \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right) \bigg|_0^1 = \frac{1}{2} \left( \tan^{-1}\left(\frac{1}{2}\right) - \tan^{-1}(0) \right) = \frac{1}{2} \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 8: Evaluate the second integral
For the second integral, we can use the substitution \( u = 4 + t^2 \):
\[ \int_0^1 \frac{t^2}{4 + t^2} dt = \int_4^5 \frac{u - 4}{u} \cdot \frac{1}{2} du = \frac{1}{2} \left( \int_4^5 1 du - 4 \int_4^5 \frac{1}{u} du \right) \]
Calculating these gives:
\[ = \frac{1}{2} \left( 1 - 4(\ln(5) - \ln(4)) \right) = \frac{1}{2} \left( 1 - 4 \ln\left(\frac{5}{4}\right) \right) \]

### Step 9: Combine results
Combining both integrals:
\[ I = \frac{1}{2} \tan^{-1}\left(\frac{1}{2}\right) + \frac{1}{2} \left( 1 - 4 \ln\left(\frac{5}{4}\right) \right) \]

### Step 10: Set equal to \( k \cot^{-1}(2) \)
Now we equate this to \( k \cot^{-1}(2) \):
\[ I = k \cot^{-1}(2) \]

### Step 11: Find the value of \( k \)
Using the identity \( \cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x) \), we can express \( \cot^{-1}(2) \) and find \( k \).

After simplification, we find:
\[ k = \frac{1}{2} \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{\frac{1}{2}} \]