If `f(x) = sin(lim_(t rarr 0)(2x)/picot^(-1) (x/t^2))`, then `int_(-(pi)/(2))^((pi)/(2))f(x)` dx is equal to (where , `x ne0`)

Options

A. `-2`
B. `-1`
C. 0
D. 2

Answer

To solve the problem, we need to evaluate the integral \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx \), where \( f(x) = \sin\left(\lim_{t \to 0} \frac{2x}{\pi \cot^{-1}\left(\frac{x}{t^2}\right)}\right) \).

### Step 1: Break the Integral
We can break the integral into two parts:
\[ I = \int_{-\frac{\pi}{2}}^{0} f(x) \, dx + \int_{0}^{\frac{\pi}{2}} f(x) \, dx \]
Let \( I_1 = \int_{-\frac{\pi}{2}}^{0} f(x) \, dx \) and \( I_2 = \int_{0}^{\frac{\pi}{2}} f(x) \, dx \).

### Step 2: Evaluate \( I_1 \)
For \( I_1 \):
\[ I_1 = \int_{-\frac{\pi}{2}}^{0} \sin\left(\lim_{t \to 0} \frac{2x}{\pi \cot^{-1}\left(\frac{x}{t^2}\right)}\right) \, dx \]
As \( t \to 0 \), \( \frac{x}{t^2} \to -\infty \) since \( x < 0 \). Therefore, \( \cot^{-1}(-\infty) = \pi \). Thus, we have:
\[ \lim_{t \to 0} \frac{2x}{\pi \cot^{-1}\left(\frac{x}{t^2}\right)} = \lim_{t \to 0} \frac{2x}{\pi \cdot \pi} = \frac{2x}{\pi^2} \]
So,
\[ f(x) = \sin\left(\frac{2x}{\pi}\right) \]
Thus,
\[ I_1 = \int_{-\frac{\pi}{2}}^{0} \sin\left(\frac{2x}{\pi}\right) \, dx \]

### Step 3: Integrate \( I_1 \)
The integral can be calculated:
\[ I_1 = -\frac{\pi}{2} \left[-\frac{1}{2} \cos\left(\frac{2x}{\pi}\right)\right]_{-\frac{\pi}{2}}^{0} \]
Calculating the limits:
\[ = -\frac{\pi}{2} \left[-\frac{1}{2} \left(\cos(0) - \cos(-1)\right)\right] \]
\[ = -\frac{\pi}{2} \left[-\frac{1}{2} (1 - (-1))\right] = -\frac{\pi}{2} \left[-\frac{1}{2} (2)\right] = -\frac{\pi}{2} \cdot -1 = \frac{\pi}{2} \]

### Step 4: Evaluate \( I_2 \)
For \( I_2 \):
\[ I_2 = \int_{0}^{\frac{\pi}{2}} \sin\left(\lim_{t \to 0} \frac{2x}{\pi \cot^{-1}\left(\frac{x}{t^2}\right)}\right) \, dx \]
As \( t \to 0 \), \( \frac{x}{t^2} \to \infty \) since \( x > 0 \). Therefore, \( \cot^{-1}(\infty) = 0 \). Thus, we have:
\[ \lim_{t \to 0} \frac{2x}{\pi \cot^{-1}\left(\frac{x}{t^2}\right)} = 0 \]
So,
\[ f(x) = \sin(0) = 0 \]
Thus,
\[ I_2 = \int_{0}^{\frac{\pi}{2}} 0 \, dx = 0 \]

### Step 5: Combine Results
Now, we combine the results:
\[ I = I_1 + I_2 = \frac{\pi}{2} + 0 = \frac{\pi}{2} \]

### Final Result
Thus, the value of the integral \( I \) is:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx = -1 \]