`int(dx)/(sqrt(cos^(3)xcos(x-alpha)))` is equal to

Options

A. `2sqrt(tanxsinalpha+cotalpha)+c`
B. `2cosecalphasqrt(tanxsinalpha+cosalpha)+c`
C. `2sqrt(sinalpha+tanxcosalpha)+c`
D. `2cosecalphasqrt(tanxcosalpha+sinalpha)+c`

Answer

To solve the integral

\[ I = \int \frac{dx}{\sqrt{\cos^3 x \cos(x - \alpha)}} \]

we will proceed step by step.

### Step 1: Rewrite the Integral
We start by rewriting the integral:

\[ I = \int \frac{dx}{\sqrt{\cos^3 x \cos(x - \alpha)}} \]

To simplify the expression under the square root, we can multiply and divide by \(\cos x\):

\[ I = \int \frac{\cos x \, dx}{\cos x \sqrt{\cos^3 x \cos(x - \alpha)}} = \int \frac{\cos x \, dx}{\sqrt{\cos^4 x \cos(x - \alpha)}} \]

### Step 2: Simplify the Square Root
Now, we can simplify the square root:

\[ I = \int \frac{\cos x \, dx}{\sqrt{\cos^4 x} \sqrt{\cos(x - \alpha)}} = \int \frac{\cos x \, dx}{\cos^2 x \sqrt{\cos(x - \alpha)}} \]

This simplifies to:

\[ I = \int \frac{dx}{\cos x \sqrt{\cos(x - \alpha)}} \]

### Step 3: Use Trigonometric Identity
Using the cosine subtraction formula, we have:

\[ \cos(x - \alpha) = \cos x \cos \alpha + \sin x \sin \alpha \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{dx}{\cos x \sqrt{\cos x \cos \alpha + \sin x \sin \alpha}} \]

### Step 4: Substitution
Let’s make the substitution:

\[ t^2 = \cos \alpha + \tan x \sin \alpha \]

Differentiating both sides gives:

\[ 2t \, dt = \sec^2 x \sin \alpha \, dx \]

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

\[ dx = \frac{2t}{\sin \alpha} dt \]

### Step 5: Substitute Back into the Integral
Substituting back into the integral, we have:

\[ I = \int \frac{2t}{\sin \alpha} \cdot \frac{1}{\sqrt{t^2}} dt = \int \frac{2t}{\sin \alpha} \cdot \frac{1}{t} dt \]

This simplifies to:

\[ I = \frac{2}{\sin \alpha} \int dt \]

### Step 6: Integrate
The integral of \(dt\) is simply \(t\):

\[ I = \frac{2}{\sin \alpha} t + C \]

### Step 7: Substitute Back for \(t\)
Now, substitute back for \(t\):

\[ t = \sqrt{\cos \alpha + \tan x \sin \alpha} \]

Thus, we have:

\[ I = \frac{2}{\sin \alpha} \sqrt{\cos \alpha + \tan x \sin \alpha} + C \]

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

\[ I = \frac{2}{\sin \alpha} \sqrt{\cos \alpha + \tan x \sin \alpha} + C \]