The value of `int(dx)/sqrt(1-x)` is

Options

A. `2sqrt(1-x)+c`
B. `-2sqrt(1-x)+c`
C. `-sin^(-1)sqrt(x)+c`
D. `sin^(-1)sqrt(x)+c`

Answer

To solve the integral \(\int \frac{dx}{\sqrt{1-x}}\), we can follow these steps:

### Step 1: Substitution
Let \( t = 1 - x \). Then, we differentiate both sides:
\[ dt = -dx \quad \Rightarrow \quad dx = -dt \]

### Step 2: Rewrite the Integral
Substituting \( t \) and \( dx \) into the integral, we get:
\[ \int \frac{dx}{\sqrt{1-x}} = \int \frac{-dt}{\sqrt{t}} = -\int \frac{dt}{\sqrt{t}} \]

### Step 3: Integrate
The integral \(-\int \frac{dt}{\sqrt{t}}\) can be solved as follows:
\[ -\int t^{-\frac{1}{2}} dt = -\left(2t^{\frac{1}{2}}\right) + C = -2\sqrt{t} + C \]

### Step 4: Back Substitute
Now, we substitute back \( t = 1 - x \):
\[ -2\sqrt{t} + C = -2\sqrt{1-x} + C \]

### Final Answer
Thus, the value of the integral is:
\[ \int \frac{dx}{\sqrt{1-x}} = -2\sqrt{1-x} + C \]

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