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

Options

A. `(-1)/(a+b)log|asin^2 x+bcos^2x|+c`
B. `(-1)/(a-b)log|asin^2 x+bcos^2x|+c`
C. `(1)/(a-b)log|asin^2 x+bcos^2x|+c`
D. `(1)/(a+b)log|asin^2 x+bcos^2x|+c`

Answer

To solve the integral

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

we can follow these steps:

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

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

Using the identity \(\sin 2x = 2 \sin x \cos x\), we can rewrite the integral as:

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

### Step 2: Substitution
Let \(t = a \sin^2 x + b \cos^2 x\). We need to find \(dt\):

\[ dt = (2a \sin x \cos x - 2b \sin x \cos x) \, dx = 2(a - b) \sin x \cos x \, dx. \]

Thus, we can express \(dx\) in terms of \(dt\):

\[ dx = \frac{dt}{2(a - b) \sin x \cos x}. \]

### Step 3: Substitute in the Integral
Substituting \(t\) and \(dx\) into the integral, we have:

\[ I = \int \frac{2 \sin x \cos x}{t} \cdot \frac{dt}{2(a - b) \sin x \cos x}. \]

The \(\sin x \cos x\) terms cancel out:

\[ I = \int \frac{1}{t} \cdot \frac{dt}{(a - b)} = \frac{1}{(a - b)} \int \frac{dt}{t}. \]

### Step 4: Integrate
The integral \(\int \frac{dt}{t}\) is:

\[ \int \frac{dt}{t} = \log |t| + C. \]

Thus, we have:

\[ I = \frac{1}{(a - b)} \log |t| + C. \]

### Step 5: Substitute Back
Now we substitute back for \(t\):

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

### Final Answer
The final result for the integral is:

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