`int(dx)/(a^2+b^2cos^2x)=`

Options

A. `(1)/(bsqrt(a^2+b^2))tan^(-1)((btanx)/(sqrt(a^2+b^2)))+c`
B. `(-1)/(bsqrt(a^2+b^2))tan^(-1)((btanx)/(sqrt(a^2+b^2)))+c`
C. `(1)/(asqrt(a^2+b^2))tan^(-1)((atanx)/(sqrt(a^2+b^2)))+c`
D. `(-1)/(asqrt(a^2+b^2))tan^(-1)((atanx)/(sqrt(a^2+b^2)))+c`

Answer

To solve the integral \( I = \int \frac{dx}{a^2 + b^2 \cos^2 x} \), we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the Integral**:
\[ I = \int \frac{dx}{a^2 + b^2 \cos^2 x} \]

2. **Multiply by Secant Squared**:
We multiply the numerator and denominator by \( \sec^2 x \):
\[ I = \int \frac{\sec^2 x \, dx}{a^2 \sec^2 x + b^2} \]

3. **Use the Identity**:
Recall that \( \sec^2 x = 1 + \tan^2 x \):
\[ I = \int \frac{\sec^2 x \, dx}{a^2 + b^2 + a^2 \tan^2 x} \]

4. **Substitution**:
Let \( t = \tan x \). Then, \( dt = \sec^2 x \, dx \):
\[ I = \int \frac{dt}{a^2 + b^2 + a^2 t^2} \]

5. **Rearranging the Denominator**:
We can rewrite the integral:
\[ I = \int \frac{dt}{a^2 + b^2 + a^2 t^2} = \int \frac{dt}{\sqrt{a^2 + b^2}^2 + a^2 t^2} \]

6. **Standard Form**:
This integral is now in the form of:
\[ \int \frac{dt}{A^2 + B^2 t^2} \]
where \( A = \sqrt{a^2 + b^2} \) and \( B = a \).

7. **Using the Formula**:
The integral can be solved using the formula:
\[ \int \frac{dx}{A^2 + B^2 x^2} = \frac{1}{AB} \tan^{-1} \left( \frac{B x}{A} \right) + C \]
Applying this:
\[ I = \frac{1}{\sqrt{a^2 + b^2} \cdot a} \tan^{-1} \left( \frac{a t}{\sqrt{a^2 + b^2}} \right) + C \]

8. **Back Substitute**:
Since \( t = \tan x \):
\[ I = \frac{1}{\sqrt{a^2 + b^2} \cdot a} \tan^{-1} \left( \frac{a \tan x}{\sqrt{a^2 + b^2}} \right) + C \]

### Final Result:
Thus, the integral evaluates to:
\[ I = \frac{1}{a \sqrt{a^2 + b^2}} \tan^{-1} \left( \frac{a \tan x}{\sqrt{a^2 + b^2}} \right) + C \]