The area bounded by the curve `y=sin^(-1)(sinx)` and the x - axis from `x=0" to "x=4pi` is equal to the area bounded by the curve `y=cos^(-1)(cosx)` and the x - axis from `x=-pi " to "x=a`, then the value of a is equal to

Options

A. `(pi)/(2)`
B. `2pi`
C. `pi`
D. `(3pi)/(2)`

Answer

To solve the problem, we need to find the area bounded by the curves \( y = \sin^{-1}(\sin x) \) from \( x = 0 \) to \( x = 4\pi \) and \( y = \cos^{-1}(\cos x) \) from \( x = -\pi \) to \( x = a \), and set these two areas equal to each other.

### Step 1: Analyze the first curve \( y = \sin^{-1}(\sin x) \)

The function \( y = \sin^{-1}(\sin x) \) oscillates between \( 0 \) and \( \pi \) for each interval of \( 2\pi \). Specifically, it can be expressed as:
- \( y = x \) for \( 0 \leq x < \pi \)
- \( y = 2\pi - x \) for \( \pi \leq x < 2\pi \)
- \( y = x - 2\pi \) for \( 2\pi \leq x < 3\pi \)
- \( y = 4\pi - x \) for \( 3\pi \leq x < 4\pi \)

### Step 2: Calculate the area from \( x = 0 \) to \( x = 4\pi \)

We can break the area into four segments:
1. From \( 0 \) to \( \pi \): Area = \( \int_0^{\pi} x \, dx = \left[ \frac{x^2}{2} \right]_0^{\pi} = \frac{\pi^2}{2} \)
2. From \( \pi \) to \( 2\pi \): Area = \( \int_{\pi}^{2\pi} (2\pi - x) \, dx = \left[ 2\pi x - \frac{x^2}{2} \right]_{\pi}^{2\pi} = 2\pi(2\pi) - \frac{(2\pi)^2}{2} - (2\pi^2 - \frac{\pi^2}{2}) = \frac{\pi^2}{2} \)
3. From \( 2\pi \) to \( 3\pi \): Area = \( \int_{2\pi}^{3\pi} (x - 2\pi) \, dx = \frac{\pi^2}{2} \)
4. From \( 3\pi \) to \( 4\pi \): Area = \( \int_{3\pi}^{4\pi} (4\pi - x) \, dx = \frac{\pi^2}{2} \)

Total area from \( 0 \) to \( 4\pi \):
\[ \text{Total Area} = 4 \times \frac{\pi^2}{2} = 2\pi^2 \]

### Step 3: Analyze the second curve \( y = \cos^{-1}(\cos x) \)

The function \( y = \cos^{-1}(\cos x) \) behaves similarly:
- \( y = x \) for \( -\pi \leq x < 0 \)
- \( y = -x \) for \( 0 \leq x < \pi \)

### Step 4: Calculate the area from \( x = -\pi \) to \( x = a \)

1. From \( -\pi \) to \( 0 \): Area = \( \int_{-\pi}^{0} (-x) \, dx = \left[ -\frac{x^2}{2} \right]_{-\pi}^{0} = \frac{\pi^2}{2} \)
2. From \( 0 \) to \( a \): Area = \( \int_{0}^{a} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{a} = \frac{a^2}{2} \)

Total area from \( -\pi \) to \( a \):
\[ \text{Total Area} = \frac{\pi^2}{2} + \frac{a^2}{2} \]

### Step 5: Set the areas equal

Now we set the two areas equal to each other:
\[ 2\pi^2 = \frac{\pi^2}{2} + \frac{a^2}{2} \]

Multiply through by 2 to eliminate the fractions:
\[ 4\pi^2 = \pi^2 + a^2 \]

Rearranging gives:
\[ a^2 = 4\pi^2 - \pi^2 = 3\pi^2 \]

Taking the square root:
\[ a = \sqrt{3}\pi \]

### Conclusion

Thus, the value of \( a \) is \( \sqrt{3}\pi \).