If `int(dx)/(cos(x-a)cos(x-b))=(1)/(A)[log((f(x))/(g(x)))]+c`, then

Options

A. `A=sin(a-b)`
B. `f(x)=sec(x-b)`
C. `g(x)=sec(x-a)`
D. `A=sin(b-a)`

Answer

To solve the integral

\[ \int \frac{dx}{\cos(x-a) \cos(x-b)}, \]

we will use a substitution method and properties of trigonometric functions. Let's go through the solution step by step.

### Step 1: Substitution
Let’s set \( t = x - a \). Then, we have:

\[ dx = dt. \]

Substituting this into the integral gives:

\[ \int \frac{dt}{\cos(t) \cos(t + (b - a))}. \]

### Step 2: Rewrite the Integral
Now, we can rewrite the cosine term:

\[ \cos(t + (b - a)) = \cos(t) \cos(b - a) - \sin(t) \sin(b - a). \]

Thus, the integral becomes:

\[ \int \frac{dt}{\cos(t) \left( \cos(t) \cos(b - a) - \sin(t) \sin(b - a) \right)}. \]

### Step 3: Simplifying the Integral
This can be simplified as:

\[ \int \frac{dt}{\cos^2(t) \cos(b - a) - \sin(t) \sin(b - a) \cos(t)}. \]

### Step 4: Factor Out Terms
Now, we can factor out \(\cos(t)\):

\[ \int \frac{dt}{\cos(t) \left( \cos(t) \cos(b - a) - \sin(t) \sin(b - a) \right)}. \]

### Step 5: Use Trigonometric Identities
We can use the identity for tangent:

\[ \frac{1}{\cos(t)} = \sec(t). \]

Thus, we can rewrite the integral as:

\[ \int \sec(t) \frac{dt}{\cos(t) \cos(b - a) - \sin(t) \sin(b - a)}. \]

### Step 6: Change of Variable
Let’s set \( v = \tan(t) \), then \( dt = \frac{dv}{\sec^2(t)} \). This gives us:

\[ \int \frac{1}{\cos(b - a) - v \sin(b - a)} dv. \]

### Step 7: Integrate
This integral can be solved using the formula for the integral of a rational function. The result will be:

\[ \frac{1}{\sin(b - a)} \ln \left| \cos(b - a) + v \sin(b - a) \right| + C. \]

### Step 8: Back Substitution
Now, we substitute back \( v = \tan(t) \) and \( t = x - a \):

\[ \frac{1}{\sin(b - a)} \ln \left| \cos(b - a) + \tan(x - a) \sin(b - a) \right| + C. \]

### Final Result
Thus, we have:

\[ \int \frac{dx}{\cos(x-a) \cos(x-b)} = \frac{1}{\sin(b - a)} \ln \left| \cos(b - a) + \tan(x - a) \sin(b - a) \right| + C. \]