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 can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral in a more manageable form. We can factor out \( \cos^2 x \) from the denominator:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{\cos^2 x \left( a^2 + b^2 \tan^2 x \right)}. \]

### Step 2: Substitute for \( \tan x \)
Next, we make the substitution \( z = \tan x \). Then, we have:

\[ dx = \frac{dz}{\sec^2 x} = \frac{dz}{1 + z^2}. \]

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

Substituting these into the integral gives:

\[ I = \int_{0}^{\infty} \frac{dz}{(a^2 + b^2 z^2)(1 + z^2)}. \]

### Step 3: Factor the Denominator
Now we can express the integral as:

\[ I = \int_{0}^{\infty} \frac{dz}{b^2 \left( \frac{a^2}{b^2} + z^2 \right)(1 + z^2)}. \]

### Step 4: Use Partial Fraction Decomposition
We can use partial fraction decomposition on the integrand:

\[ \frac{1}{\left( \frac{a^2}{b^2} + z^2 \right)(1 + z^2)} = \frac{A}{\frac{a^2}{b^2} + z^2} + \frac{B}{1 + z^2}. \]

To find \( A \) and \( B \), we multiply through by the denominator:

\[ 1 = A(1 + z^2) + B\left(\frac{a^2}{b^2} + z^2\right). \]

### Step 5: Solve for Coefficients
Setting \( z = 0 \):

\[ 1 = A + B\frac{a^2}{b^2} \implies B = \frac{1 - A}{\frac{a^2}{b^2}}. \]

Setting \( z = i \):

\[ 1 = A(1 - 1) + B\left(\frac{a^2}{b^2} - 1\right) \implies B = \frac{1}{\frac{a^2}{b^2} - 1}. \]

### Step 6: Integrate Each Term
Now we can integrate each term separately. The integral of \( \frac{1}{\frac{a^2}{b^2} + z^2} \) is:

\[ \frac{1}{\sqrt{\frac{a^2}{b^2}}} \tan^{-1}\left(\frac{z}{\sqrt{\frac{a^2}{b^2}}}\right). \]

The integral of \( \frac{1}{1 + z^2} \) is:

\[ \tan^{-1}(z). \]

### Step 7: Combine Results
Combining the results and evaluating from \( 0 \) to \( \infty \):

\[ I = \frac{1}{b^2} \left( \frac{b}{a} \cdot \frac{\pi}{2} + \frac{\pi}{2} \right) = \frac{\pi}{2ab}. \]

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

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