If `int(cosx-sinx)/(sqrt(8-sin2x))dx=sin^(-1)((sinx+cosx)/(a))+C` then a =

Options

A. 2
B. 3
C. 4
D. none of these

Explanation

To solve the integral

\[ I = \int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} \, dx \]

and find the value of \( a \) such that

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

we will follow these steps:

### Step 1: Simplify the Denominator
We start with the expression in the denominator:

\[ \sqrt{8 - \sin 2x}. \]

Recall that \( \sin 2x = 2 \sin x \cos x \). Thus, we can rewrite the expression as:

\[ 8 - \sin 2x = 8 - 2 \sin x \cos x. \]

Now, we can express it as:

\[ 8 - 2 \sin x \cos x = 8 - 2 \cdot \frac{1}{2} \cdot 2 \sin x \cos x = 8 - 2 \cdot \sin x \cos x. \]

### Step 2: Rewrite the Denominator
Next, we can add and subtract 1 to the expression:

\[ 8 - 2 \sin x \cos x = 9 - (\sin x + \cos x)^2. \]

This is because:

\[ (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + 2 \sin x \cos x. \]

Thus, we can rewrite:

\[ 8 - 2 \sin x \cos x = 9 - (\sin x + \cos x)^2. \]

### Step 3: Change of Variable
Now, let \( u = \sin x + \cos x \). Then, the differential \( du \) is:

\[ du = (\cos x - \sin x) \, dx. \]

This means we can rewrite our integral as:

\[ I = \int \frac{du}{\sqrt{9 - u^2}}. \]

### Step 4: Solve the Integral
The integral

\[ \int \frac{du}{\sqrt{9 - u^2}} \]

is a standard integral that evaluates to:

\[ \sin^{-1}\left(\frac{u}{3}\right) + C. \]

### Step 5: Substitute Back
Substituting back for \( u \):

\[ I = \sin^{-1}\left(\frac{\sin x + \cos x}{3}\right) + C. \]

### Step 6: Compare with Given Expression
We are given that:

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

By comparing both expressions, we find:

\[ a = 3. \]

### Final Answer
Thus, the value of \( a \) is

\[ \boxed{3}. \]