`int(sinx. cos x)/(a^(2) cos^(2) x+b^(2) sin^(2) x)dx`

Answer

To solve the integral

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

we can follow these steps:

### Step 1: Simplify the Denominator
We can express \(\sin^2 x\) in terms of \(\cos^2 x\) using the identity \(\sin^2 x = 1 - \cos^2 x\):

\[ I = \int \frac{\sin x \cos x}{a^2 \cos^2 x + b^2 (1 - \cos^2 x)} \, dx. \]

### Step 2: Rewrite the Denominator
Now, substitute \(\sin^2 x\):

\[ I = \int \frac{\sin x \cos x}{(a^2 - b^2) \cos^2 x + b^2} \, dx. \]

### Step 3: Substitute for \(\cos x\)
Let \(t = \cos x\), then \(dt = -\sin x \, dx\) or \(-dt = \sin x \, dx\). The integral becomes:

\[ I = -\int \frac{t}{(a^2 - b^2)t^2 + b^2} \, dt. \]

### Step 4: Factor out Constants
We can factor out the constant from the denominator:

\[ I = -\frac{1}{a^2 - b^2} \int \frac{t}{t^2 + \frac{b^2}{a^2 - b^2}} \, dt. \]

### Step 5: Use a Suitable Substitution
Let \(u = t^2 + \frac{b^2}{a^2 - b^2}\), then \(du = 2t \, dt\) or \(dt = \frac{du}{2t}\). The integral becomes:

\[ I = -\frac{1}{2(a^2 - b^2)} \int \frac{1}{u} \, du. \]

### Step 6: Integrate
The integral of \(\frac{1}{u}\) is \(\ln |u|\):

\[ I = -\frac{1}{2(a^2 - b^2)} \ln |u| + C. \]

### Step 7: Substitute Back for \(u\)
Substituting back for \(u\):

\[ I = -\frac{1}{2(a^2 - b^2)} \ln \left| t^2 + \frac{b^2}{a^2 - b^2} \right| + C. \]

### Step 8: Substitute Back for \(t\)
Now substitute \(t = \cos x\):

\[ I = -\frac{1}{2(a^2 - b^2)} \ln \left| \cos^2 x + \frac{b^2}{a^2 - b^2} \right| + C. \]

### Final Answer
Thus, the final answer for the integral is:

\[ I = -\frac{1}{2(a^2 - b^2)} \ln \left| \cos^2 x + \frac{b^2}{a^2 - b^2} \right| + C. \]

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