`int_(0)^(1)tan ^(-1) (x/sqrt(1-x^(2)))dx=`

Options

A. `pi/2-1`
B. `1-pi/2`
C. `pi/2-2`
D. `2-pi/2`

Answer

To solve the integral

\[ I = \int_{0}^{1} \tan^{-1} \left( \frac{x}{\sqrt{1-x^2}} \right) dx, \]

we can use a trigonometric substitution. Let's follow the steps outlined in the video transcript.

### Step 1: Substitution
Let \( x = \sin \theta \). Then, we differentiate both sides to find \( dx \):

\[ dx = \cos \theta \, d\theta. \]

### Step 2: Change the limits of integration
When \( x = 0 \), \( \theta = 0 \) and when \( x = 1 \), \( \theta = \frac{\pi}{2} \). Thus, the limits change from \( 0 \) to \( \frac{\pi}{2} \).

### Step 3: Substitute in the integral
Substituting \( x \) and \( dx \) into the integral gives:

\[ I = \int_{0}^{\frac{\pi}{2}} \tan^{-1} \left( \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}} \right) \cos \theta \, d\theta. \]

Since \( \sqrt{1 - \sin^2 \theta} = \cos \theta \), we can simplify the argument of the arctangent:

\[ I = \int_{0}^{\frac{\pi}{2}} \tan^{-1} \left( \frac{\sin \theta}{\cos \theta} \right) \cos \theta \, d\theta. \]

### Step 4: Simplify the arctangent
The expression \( \frac{\sin \theta}{\cos \theta} \) is simply \( \tan \theta \). Therefore, we have:

\[ I = \int_{0}^{\frac{\pi}{2}} \tan^{-1} (\tan \theta) \cos \theta \, d\theta. \]

Since \( \tan^{-1} (\tan \theta) = \theta \) for \( \theta \in [0, \frac{\pi}{2}] \), we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} \theta \cos \theta \, d\theta. \]

### Step 5: Integration by parts
Now, we apply integration by parts. Let:

- \( u = \theta \) \(\Rightarrow du = d\theta\),
- \( dv = \cos \theta \, d\theta \) \(\Rightarrow v = \sin \theta\).

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:

\[ I = \left[ \theta \sin \theta \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta. \]

### Step 6: Evaluate the boundary term
Evaluating the boundary term:

\[ \left[ \theta \sin \theta \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \cdot \sin\left(\frac{\pi}{2}\right) - 0 \cdot \sin(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \]

### Step 7: Evaluate the remaining integral
Now, we need to evaluate:

\[ \int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta = -\cos \theta \bigg|_{0}^{\frac{\pi}{2}} = -\left(0 - (-1)\right) = 1. \]

### Step 8: Combine results
Putting it all together, we have:

\[ I = \frac{\pi}{2} - 1. \]

Thus, the final result is:

\[ \int_{0}^{1} \tan^{-1} \left( \frac{x}{\sqrt{1-x^2}} \right) dx = \frac{\pi}{2} - 1. \]