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

\[ \int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} \, dx = \sin^{-1} \left( \frac{\sin x + \cos x}{a} \right) + C, \]

we need to find the value of \( a \).

### Step 1: Rewrite the Integral

We start by rewriting the integral. We know that

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

Thus, we can express the integral as:

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

### Step 2: Simplify the Denominator

Next, we simplify the denominator:

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

Now, we can rewrite the integral:

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

### Step 3: Factor the Denominator

To simplify the denominator further, we can add and subtract 1:

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

### Step 4: Use Substitution

Let \( t = \sin x + \cos x \). Then, the derivative is:

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

Thus, we can rewrite the integral in terms of \( t \):

\[ \int \frac{dt}{\sqrt{9 - t^2}}. \]

### Step 5: Solve the Integral

The integral

\[ \int \frac{dt}{\sqrt{9 - t^2}} \]

is a standard form, which evaluates to:

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

### Step 6: Substitute Back

Now substituting back \( t = \sin x + \cos x \):

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

### Step 7: Compare with Given Expression

From the original equation, we have:

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

By comparing both sides, we find that:

\[ a = 3. \]

### Final Answer

Thus, the value of \( a \) is

\[ \boxed{3}. \]