Let `J=int _(-1) ^(2) (cot ^(-1) ""(1)/(x )+ cot ^(-1) x ) dx, K =int _(-2pi) ^(7pi) (sin x)/(|sin x|)dx.` Then which of the following alternative (s) is/are correct ?

Options

A. `2J+3K=8pi`
B. `4J ^(2) +K ^(2) =26pi^(2)`
C. `2J =K=3pi`
D. `(J)/(K) =2/5`

Answer

To solve the problem, we need to evaluate the integrals \( J \) and \( K \) separately.

### Step 1: Evaluate \( J \)

We have:
\[ J = \int_{-1}^{2} \left( \cot^{-1}\left(\frac{1}{x}\right) + \cot^{-1}(x) \right) dx \]

Using the identity:
\[ \cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(x) \]
and the property:
\[ \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2} \]

We can rewrite \( J \) as:
\[ J = \int_{-1}^{2} \left( \tan^{-1}(x) + \cot^{-1}(x) \right) dx = \int_{-1}^{2} \frac{\pi}{2} dx \]

Now, we can calculate the integral:
\[ J = \frac{\pi}{2} \int_{-1}^{2} dx = \frac{\pi}{2} [x]_{-1}^{2} = \frac{\pi}{2} (2 - (-1)) = \frac{\pi}{2} \cdot 3 = \frac{3\pi}{2} \]

### Step 2: Evaluate \( K \)

Next, we evaluate:
\[ K = \int_{-2\pi}^{7\pi} \frac{\sin x}{|\sin x|} dx \]

We need to analyze the function \( \frac{\sin x}{|\sin x|} \):
- \( \frac{\sin x}{|\sin x|} = 1 \) when \( \sin x > 0 \)
- \( \frac{\sin x}{|\sin x|} = -1 \) when \( \sin x < 0 \)

Now, we can break the integral into intervals where \( \sin x \) is positive and negative.

The periodicity of \( \sin x \) is \( 2\pi \). The intervals we need to consider from \( -2\pi \) to \( 7\pi \) are:
1. From \( -2\pi \) to \( -\pi \): \( \sin x < 0 \) (contributes \(-\pi\))
2. From \( -\pi \) to \( 0 \): \( \sin x < 0 \) (contributes \(-\pi\))
3. From \( 0 \) to \( \pi \): \( \sin x > 0 \) (contributes \( \pi \))
4. From \( \pi \) to \( 2\pi \): \( \sin x < 0 \) (contributes \(-\pi\))
5. From \( 2\pi \) to \( 3\pi \): \( \sin x > 0 \) (contributes \( \pi \))
6. From \( 3\pi \) to \( 4\pi \): \( \sin x < 0 \) (contributes \(-\pi\))
7. From \( 4\pi \) to \( 5\pi \): \( \sin x > 0 \) (contributes \( \pi \))
8. From \( 5\pi \) to \( 6\pi \): \( \sin x < 0 \) (contributes \(-\pi\))
9. From \( 6\pi \) to \( 7\pi \): \( \sin x > 0 \) (contributes \( \pi \))

Now, we can sum these contributions:
- From \( -2\pi \) to \( -\pi \): \(-\pi\)
- From \( -\pi \) to \( 0 \): \(-\pi\)
- From \( 0 \) to \( \pi \): \(+\pi\)
- From \( \pi \) to \( 2\pi \): \(-\pi\)
- From \( 2\pi \) to \( 3\pi \): \(+\pi\)
- From \( 3\pi \) to \( 4\pi \): \(-\pi\)
- From \( 4\pi \) to \( 5\pi \): \(+\pi\)
- From \( 5\pi \) to \( 6\pi \): \(-\pi\)
- From \( 6\pi \) to \( 7\pi \): \(+\pi\)

Calculating the total:
\[ K = -\pi - \pi + \pi - \pi + \pi - \pi + \pi - \pi + \pi = \pi \]

### Conclusion

Now we have:
\[ J = \frac{3\pi}{2}, \quad K = \pi \]

### Step 3: Analyze the options

1. \( 2J + 3K = 2 \cdot \frac{3\pi}{2} + 3\pi = 3\pi + 3\pi = 6\pi \) (Incorrect)
2. \( 4J^2 + K^2 = 4\left(\frac{3\pi}{2}\right)^2 + \pi^2 = 4 \cdot \frac{9\pi^2}{4} + \pi^2 = 9\pi^2 + \pi^2 = 10\pi^2 \) (Incorrect)
3. \( 2J = 3K \) gives \( 2 \cdot \frac{3\pi}{2} = 3\pi \) (Correct)
4. \( \frac{J}{K} = \frac{\frac{3\pi}{2}}{\pi} = \frac{3}{2} \) (Correct)

### Final Answer
The correct alternatives are:
- 3. \( 2J = 3K \)
- 4. \( \frac{J}{K} = \frac{3}{2} \)