Evaluate the following integrals:
`int(dx)/((acosx+bsinx)^(2)),agt0andbgt0`

Explanation

To evaluate the integral

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

where \( a > 0 \) and \( b > 0 \), we can follow these steps:

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

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

### Step 2: Factor out \( b \)
We can factor out \( b \) from the denominator:

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

This simplifies to:

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

### Step 3: Substitute \( t = \frac{a}{b} \cos x + \sin x \)
Let’s make the substitution:

\[ t = \frac{a}{b} \cos x + \sin x \]

Now, we need to find \( dt \). The derivative of \( t \) with respect to \( x \) is:

\[ dt = \left(-\frac{a}{b} \sin x + \cos x\right) dx \]

This can be rewritten as:

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

### Step 4: Solve for \( dx \)
We can express \( dx \) in terms of \( dt \):

\[ dx = \frac{dt}{\cos x - \frac{a}{b} \sin x} \]

### Step 5: Substitute back into the integral
Now, substituting back into the integral, we have:

\[ I = \frac{1}{b^2} \int \frac{dx}{t^2} \]

Substituting \( dx \):

\[ I = \frac{1}{b^2} \int \frac{dt}{t^2 \left(\cos x - \frac{a}{b} \sin x\right)} \]

### Step 6: Evaluate the integral
The integral \( \int \frac{dt}{t^2} \) is:

\[ -\frac{1}{t} + C \]

### Step 7: Substitute back for \( t \)
Substituting back for \( t \):

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

### Final Result
Thus, the final result for the integral is:

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