If `int(x+(cos^(-1)3x)^(2))/(sqrt(1-9x^(2)))dx=Asqrt(1-9x^(2))+B(cos^(-1)3x)^(3)+C,` then A-B is

Options

A. `(2)/(9)`
B. `-(1)/(9)`
C. `(1)/(9)`
D. 0

Explanation

To solve the integral

\[ \int \frac{x + (\cos^{-1}(3x))^2}{\sqrt{1 - 9x^2}} \, dx = A\sqrt{1 - 9x^2} + B(\cos^{-1}(3x))^3 + C, \]

we will break the integral into two parts:

1. \( I_1 = \int \frac{x}{\sqrt{1 - 9x^2}} \, dx \)
2. \( I_2 = \int \frac{(\cos^{-1}(3x))^2}{\sqrt{1 - 9x^2}} \, dx \)

### Step 1: Solve \( I_1 \)

For \( I_1 \):

Let \( t = 1 - 9x^2 \). Then, differentiating gives:

\[ dt = -18x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{18x} \]

Substituting \( x^2 = \frac{1 - t}{9} \) gives \( x = \sqrt{\frac{1 - t}{9}} = \frac{\sqrt{1 - t}}{3} \).

Thus, we have:

\[ I_1 = \int \frac{x}{\sqrt{1 - 9x^2}} \, dx = \int \frac{\frac{\sqrt{1 - t}}{3}}{\sqrt{t}} \left(-\frac{dt}{18x}\right) \]

Substituting \( x \) in terms of \( t \):

\[ I_1 = -\frac{1}{54} \int \frac{\sqrt{1 - t}}{\sqrt{t}} \, dt \]

This integral can be solved using the substitution \( t = \sin^2(\theta) \), leading to:

\[ I_1 = -\frac{1}{54} \left( -2\sqrt{1 - t} \cdot t^{1/2} + C \right) = -\frac{1}{27}\sqrt{1 - 9x^2} + C \]

### Step 2: Solve \( I_2 \)

For \( I_2 \):

Let \( u = \cos^{-1}(3x) \). Then, \( 3x = \cos(u) \) implies \( x = \frac{\cos(u)}{3} \).

Differentiating gives:

\[ dx = -\frac{\sin(u)}{3} \, du \]

Now, we need to express \( \sqrt{1 - 9x^2} \):

\[ \sqrt{1 - 9x^2} = \sqrt{1 - 9\left(\frac{\cos(u)}{3}\right)^2} = \sqrt{1 - \cos^2(u)} = \sin(u) \]

Thus, we can rewrite \( I_2 \):

\[ I_2 = \int \frac{u^2}{\sin(u)} \left(-\frac{\sin(u)}{3} \, du\right) = -\frac{1}{3} \int u^2 \, du \]

Integrating gives:

\[ I_2 = -\frac{1}{3} \cdot \frac{u^3}{3} + C = -\frac{1}{9} (\cos^{-1}(3x))^3 + C \]

### Step 3: Combine \( I_1 \) and \( I_2 \)

Now, combining both parts:

\[ I = I_1 + I_2 = -\frac{1}{27} \sqrt{1 - 9x^2} - \frac{1}{9} (\cos^{-1}(3x))^3 + C \]

### Step 4: Identify coefficients \( A \) and \( B \)

From the expression:

\[ I = A\sqrt{1 - 9x^2} + B(\cos^{-1}(3x))^3 + C, \]

we identify:

\[ A = -\frac{1}{27}, \quad B = -\frac{1}{9} \]

### Step 5: Calculate \( A - B \)

Now, we find \( A - B \):

\[ A - B = -\frac{1}{27} - \left(-\frac{1}{9}\right) = -\frac{1}{27} + \frac{3}{27} = \frac{2}{27} \]

Thus, the final answer is:

\[ \boxed{\frac{2}{27}} \]