Evaluate: `int_0^1 sin^-1 (xsqrt(1-x) -sqrtxsqrt(1-x^2))dx.`

Answer

To evaluate the integral

\[ I = \int_0^1 \sin^{-1}(x \sqrt{1-x} - \sqrt{x} \sqrt{1-x^2}) \, dx, \]

we will simplify the integrand first.

### Step 1: Simplifying the Argument of the Inverse Sine

Let \( x = \sin a \). Then, we have:

\[ 1 - x^2 = 1 - \sin^2 a = \cos^2 a \quad \text{and} \quad \sqrt{1 - x^2} = \cos a. \]

Also, we can express \( \sqrt{1-x} \):

\[ \sqrt{1 - x} = \sqrt{1 - \sin a} = \sqrt{\cos^2 a} = \cos a. \]

Now substituting these into the integrand:

\[ x \sqrt{1-x} = \sin a \cos a, \]
and
\[ \sqrt{x} \sqrt{1-x^2} = \sqrt{\sin a} \cdot \cos a = \sin^{1/2} a \cdot \cos a. \]

Thus, the argument of the inverse sine becomes:

\[ \sin a \cos a - \sin^{1/2} a \cos a = \cos a \left( \sin a - \sin^{1/2} a \right). \]

### Step 2: Using the Inverse Sine Identity

Now we can rewrite the integral as:

\[ I = \int_0^1 \sin^{-1} \left( \cos a \left( \sin a - \sin^{1/2} a \right) \right) \, dx. \]

### Step 3: Change of Variables

Next, we can change the variable \( x \) to \( \sin a \):

\[ dx = \cos a \, da. \]

When \( x = 0 \), \( a = 0 \) and when \( x = 1 \), \( a = \frac{\pi}{2} \). Thus, the limits of integration change from \( 0 \) to \( \frac{\pi}{2} \).

### Step 4: Splitting the Integral

Now, we split the integral into two parts:

\[ I = \int_0^{\frac{\pi}{2}} \sin^{-1}(\sin a) \cos a \, da - \int_0^{\frac{\pi}{2}} \sin^{-1}(\sin^{1/2} a) \cos a \, da. \]

Let:

\[ I_1 = \int_0^{\frac{\pi}{2}} a \cos a \, da, \]
\[ I_2 = \int_0^{\frac{\pi}{2}} \sin^{-1}(\sin^{1/2} a) \cos a \, da. \]

### Step 5: Evaluating \( I_1 \)

Using integration by parts for \( I_1 \):

Let \( u = a \) and \( dv = \cos a \, da \). Then \( du = da \) and \( v = \sin a \).

\[ I_1 = a \sin a \bigg|_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \sin a \, da. \]

Calculating the limits:

\[ = \left( \frac{\pi}{2} \cdot 1 - 0 \right) - \left( -\cos a \bigg|_0^{\frac{\pi}{2}} \right) = \frac{\pi}{2} - (0 + 1) = \frac{\pi}{2} - 1. \]

### Step 6: Evaluating \( I_2 \)

For \( I_2 \):

Using the substitution \( u = \sin^{1/2} a \), we can evaluate it similarly. However, we can also note that:

\[ \sin^{-1}(\sin^{1/2} a) = \frac{1}{2} a \quad \text{(for small angles)}. \]

Thus, we can evaluate:

\[ I_2 = \int_0^{\frac{\pi}{2}} \frac{1}{2} a \cos a \, da = \frac{1}{2} I_1. \]

### Step 7: Final Calculation

Putting it all together:

\[ I = I_1 - I_2 = \left( \frac{\pi}{2} - 1 \right) - \frac{1}{2} \left( \frac{\pi}{2} - 1 \right) = \frac{1}{2} \left( \frac{\pi}{2} - 1 \right). \]

### Final Result

Thus, the value of the integral is:

\[ \boxed{\frac{\pi}{4} - 1}. \]