`int_(0)^(1) x^(5)sin^(-1)xdx`

Answer

To solve the integral \( I = \int_{0}^{1} x^5 \sin^{-1}(x) \, dx \), we will use integration by parts. The formula for integration by parts is:

\[ \int u \, dv = uv - \int v \, du \]

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = \sin^{-1}(x) \) (thus \( du = \frac{1}{\sqrt{1-x^2}} \, dx \))
- \( dv = x^5 \, dx \) (thus \( v = \frac{x^6}{6} \))

### Step 2: Apply the integration by parts formula

Using the integration by parts formula, we have:

\[ I = \left[ \sin^{-1}(x) \cdot \frac{x^6}{6} \right]_{0}^{1} - \int_{0}^{1} \frac{x^6}{6} \cdot \frac{1}{\sqrt{1-x^2}} \, dx \]

### Step 3: Evaluate the boundary term

Now, we evaluate the boundary term:

\[ \left[ \sin^{-1}(x) \cdot \frac{x^6}{6} \right]_{0}^{1} = \left( \sin^{-1}(1) \cdot \frac{1^6}{6} \right) - \left( \sin^{-1}(0) \cdot \frac{0^6}{6} \right) \]

Calculating this gives:

\[ = \left( \frac{\pi}{2} \cdot \frac{1}{6} \right) - (0) = \frac{\pi}{12} \]

### Step 4: Simplify the integral

Now we need to simplify the integral:

\[ I = \frac{\pi}{12} - \frac{1}{6} \int_{0}^{1} \frac{x^6}{\sqrt{1-x^2}} \, dx \]

### Step 5: Solve the integral \( \int_{0}^{1} \frac{x^6}{\sqrt{1-x^2}} \, dx \)

To solve this integral, we can use the substitution \( x = \sin(t) \), which gives \( dx = \cos(t) \, dt \) and changes the limits from \( 0 \) to \( \frac{\pi}{2} \):

\[ \int_{0}^{1} \frac{x^6}{\sqrt{1-x^2}} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin^6(t)}{\sqrt{1-\sin^2(t)}} \cos(t) \, dt = \int_{0}^{\frac{\pi}{2}} \sin^6(t) \, dt \]

### Step 6: Evaluate \( \int_{0}^{\frac{\pi}{2}} \sin^6(t) \, dt \)

Using the reduction formula or known results, we find:

\[ \int_{0}^{\frac{\pi}{2}} \sin^n(t) \, dt = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2} \]

For \( n = 6 \):

\[ \int_{0}^{\frac{\pi}{2}} \sin^6(t) \, dt = \frac{5!!}{6!!} \cdot \frac{\pi}{2} = \frac{15}{48} \cdot \frac{\pi}{2} = \frac{15\pi}{96} \]

### Step 7: Substitute back into the expression for \( I \)

Now substituting this back into our expression for \( I \):

\[ I = \frac{\pi}{12} - \frac{1}{6} \cdot \frac{15\pi}{96} \]

Calculating the second term:

\[ = \frac{\pi}{12} - \frac{15\pi}{576} = \frac{48\pi}{576} - \frac{15\pi}{576} = \frac{33\pi}{576} = \frac{11\pi}{192} \]

### Final Result

Thus, the value of the integral is:

\[ \boxed{\frac{11\pi}{192}} \]