`int _(0)^(1) (tan ^(-1)x)/(x ) dx =`

Options

A. `int _(0)^(pi//4) (sin x)/(x )dx `
B. `int _(0)^(pi//2) (x)/(sin x )dx `
C. `1/2 int _(0)^(pi//2) (x)/(sin x) dx`
D. `1/2 int _(0)^(pi//2) (x )/(sin x ) dx`

Answer

To solve the integral \( I = \int_0^1 \frac{\tan^{-1} x}{x} \, dx \), we can follow these steps:

### Step 1: Set up the integral
Let
\[ I = \int_0^1 \frac{\tan^{-1} x}{x} \, dx \]

### Step 2: Use substitution
We will use the substitution \( x = \tan \theta \). Then, we differentiate both sides:
\[ dx = \sec^2 \theta \, d\theta \]

### Step 3: Change the limits of integration
When \( x = 0 \):
\[ \tan \theta = 0 \implies \theta = 0 \]
When \( x = 1 \):
\[ \tan \theta = 1 \implies \theta = \frac{\pi}{4} \]
Thus, the limits change from \( x: 0 \to 1 \) to \( \theta: 0 \to \frac{\pi}{4} \).

### Step 4: Rewrite the integral
Substituting \( x = \tan \theta \) into the integral gives:
\[ I = \int_0^{\frac{\pi}{4}} \frac{\tan^{-1}(\tan \theta)}{\tan \theta} \cdot \sec^2 \theta \, d\theta \]
Since \( \tan^{-1}(\tan \theta) = \theta \) for \( \theta \) in the range \( [0, \frac{\pi}{4}] \), we have:
\[ I = \int_0^{\frac{\pi}{4}} \frac{\theta}{\tan \theta} \cdot \sec^2 \theta \, d\theta \]

### Step 5: Simplify the integrand
Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \):
\[ I = \int_0^{\frac{\pi}{4}} \theta \cdot \frac{\sec^2 \theta}{\tan \theta} \, d\theta = \int_0^{\frac{\pi}{4}} \theta \cdot \frac{1}{\sin \theta \cos \theta} \, d\theta \]

### Step 6: Use a trigonometric identity
We can use the identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \):
\[ I = \int_0^{\frac{\pi}{4}} \frac{2\theta}{\sin(2\theta)} \, d\theta \]

### Step 7: Change of variable
Let \( t = 2\theta \), then \( d\theta = \frac{dt}{2} \). The limits change as follows:
When \( \theta = 0 \), \( t = 0 \); when \( \theta = \frac{\pi}{4} \), \( t = \frac{\pi}{2} \):
\[ I = \int_0^{\frac{\pi}{2}} \frac{t}{\sin t} \cdot \frac{1}{2} \, dt = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{t}{\sin t} \, dt \]

### Step 8: Recognize the integral
The integral \( \int_0^{\frac{\pi}{2}} \frac{t}{\sin t} \, dt \) is a known integral and evaluates to \( \frac{\pi^2}{4} \):
\[ I = \frac{1}{2} \cdot \frac{\pi^2}{4} = \frac{\pi^2}{8} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^1 \frac{\tan^{-1} x}{x} \, dx = \frac{\pi^2}{8} \]