Evaluate: `int cos^(-1) sqrt((x)/(x+1))dx`

Answer

To evaluate the integral \( \int \cos^{-1} \left( \sqrt{\frac{x}{x+1}} \right) dx \), we will use a trigonometric substitution. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( x = \cot^2 \theta \). Then, we differentiate to find \( dx \):
\[ dx = 2 \cot \theta (-\csc^2 \theta) d\theta = -2 \cot \theta \csc^2 \theta d\theta \]

### Step 2: Simplifying the Integral
Now, substituting \( x \) into the integral:
\[ \sqrt{\frac{x}{x+1}} = \sqrt{\frac{\cot^2 \theta}{\cot^2 \theta + 1}} = \sqrt{\frac{\cot^2 \theta}{\csc^2 \theta}} = \sqrt{\cos^2 \theta} = \cos \theta \]
Thus, we have:
\[ \cos^{-1} \left( \sqrt{\frac{x}{x+1}} \right) = \cos^{-1}(\cos \theta) = \theta \]

### Step 3: Substituting Back into the Integral
Now, substituting everything back into the integral:
\[ \int \cos^{-1} \left( \sqrt{\frac{x}{x+1}} \right) dx = \int \theta (-2 \cot \theta \csc^2 \theta) d\theta \]
This simplifies to:
\[ -2 \int \theta \cot \theta \csc^2 \theta d\theta \]

### Step 4: Integration by Parts
Using integration by parts, let:
- \( u = \theta \) (thus \( du = d\theta \))
- \( dv = \cot \theta \csc^2 \theta d\theta \)

Now, we need to find \( v \):
\[ v = \int \cot \theta \csc^2 \theta d\theta \]
Using the identity \( \frac{d}{d\theta}(\cot \theta) = -\csc^2 \theta \), we find:
\[ v = -\frac{1}{2} \cot^2 \theta \]

### Step 5: Applying Integration by Parts Formula
Now applying the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ -2 \left( \theta \left(-\frac{1}{2} \cot^2 \theta\right) - \int \left(-\frac{1}{2} \cot^2 \theta\right) d\theta \right) \]
This simplifies to:
\[ \cot^2 \theta \theta + \int \frac{1}{2} \cot^2 \theta d\theta \]

### Step 6: Evaluating the Remaining Integral
The integral \( \int \cot^2 \theta d\theta \) can be evaluated using the identity \( \cot^2 \theta = \csc^2 \theta - 1 \):
\[ \int \cot^2 \theta d\theta = \int (\csc^2 \theta - 1) d\theta = -\cot \theta - \theta \]
Thus, we have:
\[ \int \frac{1}{2} \cot^2 \theta d\theta = -\frac{1}{2} \cot \theta - \frac{1}{2} \theta \]

### Step 7: Final Expression
Combining everything, we get:
\[ \cot^2 \theta \theta + \left(-\frac{1}{2} \cot \theta - \frac{1}{2} \theta\right) \]
Substituting back \( \theta = \cot^{-1}(\sqrt{x}) \) and \( \cot \theta = \sqrt{x} \):
\[ = \sqrt{x}^2 \cot^{-1}(\sqrt{x}) - \frac{1}{2} \sqrt{x} - \frac{1}{2} \cot^{-1}(\sqrt{x}) + C \]
Thus, the final result is:
\[ \int \cos^{-1} \left( \sqrt{\frac{x}{x+1}} \right) dx = x \cot^{-1}(\sqrt{x}) - \frac{1}{2} \sqrt{x} - \frac{1}{2} \cot^{-1}(\sqrt{x}) + C \]