`int(cotx)/(sqrt(sinx))dx=`

Options

A. `-(2)/(sqrt(sinx))+c`
B. `(2)/(sqrt(sinx)+c`
C. `2sqrt(sinx)+c`
D. `(1)/(2sqrt(sinx))+c`

Answer

To solve the integral \( \int \frac{\cot x}{\sqrt{\sin x}} \, dx \), we will follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting \( \cot x \) in terms of sine and cosine:
\[ \cot x = \frac{\cos x}{\sin x} \]
Thus, we can rewrite the integral as:
\[ \int \frac{\cot x}{\sqrt{\sin x}} \, dx = \int \frac{\cos x}{\sin x \sqrt{\sin x}} \, dx \]

### Step 2: Substitution
Next, we will use the substitution \( \sin x = t \). Therefore, the differential \( dx \) can be expressed in terms of \( dt \):
\[ \frac{d}{dx}(\sin x) = \cos x \implies dx = \frac{dt}{\cos x} \]
Substituting \( \sin x = t \) into the integral gives:
\[ \int \frac{\cos x}{t \sqrt{t}} \cdot \frac{dt}{\cos x} = \int \frac{1}{t \sqrt{t}} \, dt \]

### Step 3: Simplify the integrand
We can simplify \( \frac{1}{t \sqrt{t}} \) as follows:
\[ \frac{1}{t \sqrt{t}} = \frac{1}{t^{3/2}} \]
Thus, the integral now becomes:
\[ \int t^{-3/2} \, dt \]

### Step 4: Integrate
We use the power rule for integration:
\[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \]
In our case, \( n = -\frac{3}{2} \):
\[ \int t^{-3/2} \, dt = \frac{t^{-3/2 + 1}}{-3/2 + 1} + C = \frac{t^{-1/2}}{-1/2} + C = -2t^{-1/2} + C \]

### Step 5: Substitute back
Now we substitute back \( t = \sin x \):
\[ -2(\sin x)^{-1/2} + C = -\frac{2}{\sqrt{\sin x}} + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{\cot x}{\sqrt{\sin x}} \, dx = -\frac{2}{\sqrt{\sin x}} + C \]

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