Evaluate the following : `int_(0)^(pi//2)(dx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)`

Explanation

To evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \]

we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \]

### Step 2: Divide by \(\cos^2 x\)
We can simplify the integrand by dividing both the numerator and the denominator by \(\cos^2 x\):

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{a^2 + b^2 \tan^2 x} \]

### Step 3: Substitute \(t = \tan x\)
Let \(t = \tan x\). Then, the differential \(dx\) can be expressed as:

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

The limits of integration change as follows:
- When \(x = 0\), \(t = \tan(0) = 0\)
- When \(x = \frac{\pi}{2}\), \(t = \tan\left(\frac{\pi}{2}\right) = \infty\)

Substituting these into the integral gives:

\[ I = \int_{0}^{\infty} \frac{1}{a^2 + b^2 t^2} \, dt \]

### Step 4: Factor out \(b^2\)
We can factor \(b^2\) out of the denominator:

\[ I = \frac{1}{b^2} \int_{0}^{\infty} \frac{1}{\frac{a^2}{b^2} + t^2} \, dt \]

### Step 5: Use the Standard Integral Formula
The integral

\[ \int_{0}^{\infty} \frac{1}{A + t^2} \, dt = \frac{\pi}{2\sqrt{A}} \]

can be applied here, where \(A = \frac{a^2}{b^2}\):

\[ I = \frac{1}{b^2} \cdot \frac{\pi}{2\sqrt{\frac{a^2}{b^2}}} = \frac{1}{b^2} \cdot \frac{\pi}{2 \cdot \frac{a}{b}} = \frac{\pi}{2ab} \]

### Final Result
Thus, the value of the integral is:

\[ I = \frac{\pi}{2ab} \]