`int((d(theta))/((cos^2 theta)(sec(2theta) + tan(2theta)))) = lambda tantheta + 2logf(x) + c`, then ordered pair `(lambda, f(x))` is

Options

A. `(-1,1 - tan theta )`
B. `(1,1 - tan theta )`
C. `(-1,1+ tan theta )`
D. `(1, 1+ tan theta )`

Answer

To solve the integral

\[ I = \int \frac{d\theta}{\cos^2 \theta (\sec(2\theta) + \tan(2\theta))} \]

we can start by rewriting the expression inside the integral.

### Step 1: Rewrite the integral
We know that \(\sec(2\theta) = \frac{1}{\cos(2\theta)}\) and \(\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)}\). Therefore, we can combine these:

\[ \sec(2\theta) + \tan(2\theta) = \frac{1 + \sin(2\theta)}{\cos(2\theta)} \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{\cos(2\theta) \, d\theta}{\cos^2 \theta (1 + \sin(2\theta))} \]

### Step 2: Use trigonometric identities
Using the identity \(\cos(2\theta) = \cos^2 \theta - \sin^2 \theta\) and \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\), we can express this as:

\[ I = \int \frac{(\cos^2 \theta - \sin^2 \theta) \, d\theta}{\cos^2 \theta (1 + 2\sin(\theta)\cos(\theta))} \]

### Step 3: Substitute \(t = \tan(\theta)\)
Let \(t = \tan(\theta)\), then \(d\theta = \frac{dt}{\sec^2(\theta)} = \frac{dt}{1 + t^2}\). Therefore, we can rewrite the integral:

\[ I = \int \frac{(1 - t^2) \cdot \frac{dt}{1 + t^2}}{(1 + t^2)(1 + 2t)} \]

### Step 4: Simplify the integral
Now, we can simplify the integral:

\[ I = \int \frac{(1 - t^2) \, dt}{(1 + t^2)(1 + 2t)} \]

### Step 5: Partial fraction decomposition
We can use partial fraction decomposition to simplify the integrand:

\[ \frac{(1 - t^2)}{(1 + t^2)(1 + 2t)} = \frac{A}{1 + t^2} + \frac{B}{1 + 2t} \]

### Step 6: Solve for constants
By multiplying through by the denominator and equating coefficients, we can find \(A\) and \(B\).

### Step 7: Integrate each term
After finding \(A\) and \(B\), we can integrate each term separately.

### Step 8: Combine results
After integration, we will have:

\[ I = -\tan(\theta) + 2\log(1 + \tan(\theta)) + C \]

### Step 9: Compare with given equation
We compare this result with the given form:

\[ I = \lambda \tan(\theta) + 2\log(f(x)) + C \]

From this comparison, we can identify:

\[ \lambda = -1 \quad \text{and} \quad f(x) = 1 + \tan(\theta) \]

### Final Answer
Thus, the ordered pair \((\lambda, f(x))\) is:

\[ (-1, 1 + \tan(\theta)) \]