`int(dx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)(a,bgt0)` is equal to

Options

A. `sin^(-1)((a//b)tanx) + C`
B. `tan^(-1)((b//a)tanx) + C`
C. `(1)/(ab)tan^(-1)(((b)/(a))tanx)+C`
D. none of these

Answer

To solve the integral

\[ \int \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \]

where \( a, b > 0 \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral:

\[ \int \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \]

We divide both the numerator and the denominator by \(\cos^2 x\):

\[ = \int \frac{1/\cos^2 x}{\frac{a^2 \cos^2 x}{\cos^2 x} + \frac{b^2 \sin^2 x}{\cos^2 x}} \, dx \]

This simplifies to:

\[ = \int \frac{\sec^2 x}{a^2 + b^2 \tan^2 x} \, dx \]

### Step 2: Substitute \( t = \tan x \)
Next, we use the substitution \( t = \tan x \). Then, the differential \( dx \) becomes:

\[ dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2} \]

Substituting this into the integral gives:

\[ = \int \frac{\sec^2 x}{a^2 + b^2 t^2} \cdot \frac{dt}{1 + t^2} \]

### Step 3: Simplify the Integral
Now, we can rewrite the integral as:

\[ = \int \frac{dt}{(a^2 + b^2 t^2)(1 + t^2)} \]

### Step 4: Use Partial Fraction Decomposition
To integrate this, we can use partial fraction decomposition. We express:

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

Multiplying through by the denominator:

\[ 1 = A(1 + t^2) + B(a^2 + b^2 t^2) \]

### Step 5: Solve for Constants
Setting coefficients for \( t^2 \) and constant terms, we can solve for \( A \) and \( B \).

### Step 6: Integrate Each Term
After finding \( A \) and \( B \), we integrate each term separately. The integral of \( \frac{1}{a^2 + b^2 t^2} \) is:

\[ \frac{1}{b} \tan^{-1}\left(\frac{bt}{a}\right) \]

And the integral of \( \frac{1}{1 + t^2} \) is:

\[ \tan^{-1}(t) \]

### Step 7: Combine Results
Combining the results from the integrals, we substitute back \( t = \tan x \):

\[ \int \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \frac{1}{ab} \tan^{-1}\left(\frac{b \tan x}{a}\right) + C \]

### Final Result
Thus, the final result is:

\[ \int \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \frac{1}{ab} \tan^{-1}\left(\frac{b \tan x}{a}\right) + C \]