If `int(2x-sqrt(sin^(-1)x))/(sqrt(1-x^(2)))dx=C-2sqrt(1-x^(2))-(2)/(3)sqrt(f(x))` then f(x) is equal to

Options

A. `sin^(-1)x`
B. `2sin^(-1)x`
C. `(sin^(-1)x)^(3)`
D. `3(sin^(-1)x)^(3)`

Answer

To solve the problem, we need to evaluate the integral given and find the function \( f(x) \) such that:

\[ \int \frac{2x - \sqrt{\sin^{-1} x}}{\sqrt{1 - x^2}} \, dx = C - 2\sqrt{1 - x^2} - \frac{2}{3}\sqrt{f(x)} \]

### Step-by-step Solution:

1. **Separate the Integral**:
We can separate the integral into two parts:
\[ \int \frac{2x}{\sqrt{1 - x^2}} \, dx - \int \frac{\sqrt{\sin^{-1} x}}{\sqrt{1 - x^2}} \, dx \]
Let \( I = \int \frac{2x}{\sqrt{1 - x^2}} \, dx - \int \frac{\sqrt{\sin^{-1} x}}{\sqrt{1 - x^2}} \, dx \).

2. **Evaluate the First Integral**:
For the first integral, we can use the substitution \( t = \sqrt{1 - x^2} \). Then, \( dt = -\frac{x}{\sqrt{1 - x^2}} \, dx \) or \( dx = -\frac{dt}{\sqrt{1 - x^2}} \).
The integral becomes:
\[ I_1 = \int \frac{2x}{\sqrt{1 - x^2}} \, dx = -2\sqrt{1 - x^2} + C_1 \]

3. **Evaluate the Second Integral**:
For the second integral, we can use the substitution \( u = \sin^{-1} x \), which gives \( x = \sin u \) and \( dx = \cos u \, du \). The integral becomes:
\[ I_2 = \int \frac{\sqrt{u}}{\sqrt{1 - \sin^2 u}} \cos u \, du = \int \sqrt{u} \, du \]
This evaluates to:
\[ I_2 = \frac{2}{3} u^{3/2} + C_2 = \frac{2}{3} (\sin^{-1} x)^{3/2} + C_2 \]

4. **Combine the Results**:
Now we combine \( I_1 \) and \( I_2 \):
\[ I = -2\sqrt{1 - x^2} - \frac{2}{3} (\sin^{-1} x)^{3/2} + C \]

5. **Match with Given Equation**:
We compare this with the original equation:
\[ C - 2\sqrt{1 - x^2} - \frac{2}{3}\sqrt{f(x)} \]
From this, we can identify:
\[ \sqrt{f(x)} = (\sin^{-1} x)^{3/2} \]

6. **Square Both Sides**:
To find \( f(x) \), we square both sides:
\[ f(x) = (\sin^{-1} x)^3 \]

### Final Answer:
Thus, the function \( f(x) \) is:
\[ f(x) = (\sin^{-1} x)^3 \]