Evaluate
`int_(0)^(pi//3)|(sin^(-1)sinx)/(cos^(-1)cosx)|dx`

Answer

To evaluate the integral

\[ \int_{0}^{\frac{\pi}{3}} \frac{\sin^{-1}(\sin x)}{\cos^{-1}(\cos x)} \, dx, \]

we will follow these steps:

### Step 1: Simplify the integrand
The integrand consists of \(\sin^{-1}(\sin x)\) and \(\cos^{-1}(\cos x)\). We know that:

- For \(0 \leq x \leq \frac{\pi}{2}\), \(\sin^{-1}(\sin x) = x\).
- For \(0 \leq x \leq \frac{\pi}{2}\), \(\cos^{-1}(\cos x) = x\).

Since our limits of integration are from \(0\) to \(\frac{\pi}{3}\), both \(x\) values fall within these ranges. Thus, we can simplify the integrand:

\[ \frac{\sin^{-1}(\sin x)}{\cos^{-1}(\cos x)} = \frac{x}{x} = 1. \]

### Step 2: Rewrite the integral
Now we can rewrite the integral as:

\[ \int_{0}^{\frac{\pi}{3}} 1 \, dx. \]

### Step 3: Evaluate the integral
The integral of \(1\) with respect to \(x\) is simply \(x\). Therefore, we evaluate it from \(0\) to \(\frac{\pi}{3}\):

\[ \int_{0}^{\frac{\pi}{3}} 1 \, dx = x \bigg|_{0}^{\frac{\pi}{3}} = \frac{\pi}{3} - 0 = \frac{\pi}{3}. \]

### Final Answer
Thus, the value of the integral is:

\[ \frac{\pi}{3}. \]

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