`int_(0)^(pi//2)(tanx)/(1+m^(2)tan^(2)x)\ dx`

Explanation

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\tan x}{1 + m^2 \tan^2 x} \, dx, \]

we will follow these steps:

### Step 1: Rewrite \(\tan x\)
We know that \(\tan x = \frac{\sin x}{\cos x}\). Therefore, we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\frac{\sin x}{\cos x}}{1 + m^2 \left(\frac{\sin^2 x}{\cos^2 x}\right)} \, dx. \]

### Step 2: Simplify the integrand
This can be simplified further:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\cos x} \cdot \frac{\cos^2 x}{\cos^2 x + m^2 \sin^2 x} \, dx. \]

### Step 3: Factor out \(\cos^2 x\)
Now, we can factor out \(\cos^2 x\) in the denominator:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{\cos^2 x + m^2 \sin^2 x} \, dx. \]

### Step 4: Substitute \(u = \sin^2 x\)
Let \(u = \sin^2 x\). Then, \(du = 2 \sin x \cos x \, dx\) or \(\sin x \cos x \, dx = \frac{du}{2}\). The limits change as follows: when \(x = 0\), \(u = 0\); when \(x = \frac{\pi}{2}\), \(u = 1\). Thus, we can rewrite the integral:

\[ I = \frac{1}{2} \int_{0}^{1} \frac{du}{1 - u + m^2 u}. \]

### Step 5: Simplify the integrand
The integrand simplifies to:

\[ I = \frac{1}{2} \int_{0}^{1} \frac{du}{(1 + (m^2 - 1)u)}. \]

### Step 6: Integrate
This integral can be solved using the formula for the integral of \(\frac{1}{a + bu}\):

\[ \int \frac{du}{a + bu} = \frac{1}{b} \ln |a + bu| + C. \]

Here, \(a = 1\) and \(b = m^2 - 1\):

\[ I = \frac{1}{2} \cdot \frac{1}{m^2 - 1} \left[ \ln |1 + (m^2 - 1)u| \right]_{0}^{1}. \]

### Step 7: Evaluate the limits
Now, substituting the limits:

\[ I = \frac{1}{2(m^2 - 1)} \left( \ln |1 + (m^2 - 1) \cdot 1| - \ln |1 + (m^2 - 1) \cdot 0| \right). \]

This simplifies to:

\[ I = \frac{1}{2(m^2 - 1)} \left( \ln |m^2| - \ln |1| \right). \]

Since \(\ln |1| = 0\):

\[ I = \frac{1}{2(m^2 - 1)} \ln |m^2|. \]

### Final Result
Thus, we have:

\[ I = \frac{\ln m^2}{2(m^2 - 1)} = \frac{\ln m}{m^2 - 1}. \]