The value of `int_(0)^(pi//2) (cos3x+1)/(2 cos x-1) dx ` is

Options

A. 2
B. 1
C. `1//2`
D. 0

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos 3x + 1}{2 \cos x - 1} \, dx, \]

we can start by simplifying the expression inside the integral.

### Step 1: Simplifying \(\cos 3x\)

Using the identity for \(\cos 3x\):

\[ \cos 3x = 4 \cos^3 x - 3 \cos x, \]

we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{(4 \cos^3 x - 3 \cos x) + 1}{2 \cos x - 1} \, dx. \]

### Step 2: Combine the terms in the numerator

Now, we can combine the terms in the numerator:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{4 \cos^3 x - 3 \cos x + 1}{2 \cos x - 1} \, dx. \]

### Step 3: Substitute \(1\) using \(\cos 2x\)

Recall that:

\[ 1 = 2 \cos^2 x - \cos 2x. \]

We can substitute this into our integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{4 \cos^3 x - 3 \cos x + (2 \cos^2 x - \cos 2x)}{2 \cos x - 1} \, dx. \]

### Step 4: Rearranging the numerator

Now, rearranging the numerator gives us:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{4 \cos^3 x + 2 \cos^2 x - 4 \cos x - \cos 2x}{2 \cos x - 1} \, dx. \]

### Step 5: Factor out common terms

We can factor out \(2 \cos x - 1\) from the numerator:

\[ I = \int_{0}^{\frac{\pi}{2}} \left(2 \cos x + \cos 2x\right) \, dx. \]

### Step 6: Split the integral

Now, we can split the integral into two parts:

\[ I = \int_{0}^{\frac{\pi}{2}} 2 \cos x \, dx + \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx. \]

### Step 7: Evaluate the integrals

1. For the first integral:

\[ \int 2 \cos x \, dx = 2 \sin x \bigg|_{0}^{\frac{\pi}{2}} = 2(1 - 0) = 2. \]

2. For the second integral:

\[ \int \cos 2x \, dx = \frac{1}{2} \sin 2x \bigg|_{0}^{\frac{\pi}{2}} = \frac{1}{2}(0 - 0) = 0. \]

### Step 8: Combine the results

Thus, we have:

\[ I = 2 + 0 = 2. \]

### Final Answer

Therefore, the value of the integral is:

\[ \boxed{2}. \]