If `int(1)/(4sin^(2)x+4sin x cos x+5cos^(2)x)dx=A tan^(-1)(B tanx+C)+k`, then

Options

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

Answer

To solve the integral

\[ \int \frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \, dx, \]

we will simplify the expression in the denominator and then perform the integration.

### Step 1: Simplify the Denominator

The denominator can be rewritten as:
\[ 4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x. \]

We can group and rearrange the terms:
\[ = 4 \sin^2 x + 4 \sin x \cos x + 4 \cos^2 x + \cos^2 x = 4(\sin^2 x + \sin x \cos x + \cos^2 x) + \cos^2 x. \]

Using the identity \(\sin^2 x + \cos^2 x = 1\), we have:
\[ = 4(1 + \sin x \cos x) + \cos^2 x = 4 + 4 \sin x \cos x + \cos^2 x. \]

### Step 2: Use Trigonometric Identities

Notice that:
\[ \sin x \cos x = \frac{1}{2} \sin(2x). \]
Thus, we can rewrite the denominator as:
\[ = 4 + 2 \sin(2x) + \cos^2 x. \]

### Step 3: Rewrite the Integral

Now, we can rewrite the integral:
\[ \int \frac{1}{4 + 2 \sin(2x) + \cos^2 x} \, dx. \]

### Step 4: Substitute

To simplify further, we can divide the entire expression by \(\cos^2 x\):
\[ = \int \frac{\sec^2 x}{4 \tan^2 x + 4 \tan x + 5} \, dx. \]

Let \(t = \tan x\), then \(dt = \sec^2 x \, dx\), and thus:
\[ dx = \frac{dt}{\sec^2 x}. \]

The integral becomes:
\[ \int \frac{1}{4t^2 + 4t + 5} \, dt. \]

### Step 5: Completing the Square

Now, we complete the square in the denominator:
\[ 4t^2 + 4t + 5 = 4(t^2 + t) + 5 = 4\left(t^2 + t + \frac{1}{4}\right) + 5 - 4 \cdot \frac{1}{4} = 4\left(t + \frac{1}{2}\right)^2 + 4. \]

### Step 6: Rewrite the Integral

The integral now looks like:
\[ \int \frac{1}{4\left(t + \frac{1}{2}\right)^2 + 4} \, dt = \frac{1}{4} \int \frac{1}{\left(t + \frac{1}{2}\right)^2 + 1} \, dt. \]

### Step 7: Use the Arctangent Integral Formula

Using the formula \(\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\):
\[ = \frac{1}{4} \cdot \frac{1}{1} \tan^{-1}\left(t + \frac{1}{2}\right) + C = \frac{1}{4} \tan^{-1}\left(t + \frac{1}{2}\right) + C. \]

### Step 8: Substitute Back for \(t\)

Substituting back \(t = \tan x\):
\[ = \frac{1}{4} \tan^{-1}\left(\tan x + \frac{1}{2}\right) + C. \]

### Final Result

Thus, we have:

\[ \int \frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \, dx = \frac{1}{4} \tan^{-1}\left(2 \tan x + 1\right) + C. \]

### Comparison

From the original problem, we can compare:
\[ A = \frac{1}{4}, \quad B = 2, \quad C = 1. \]