Evaluate `int_0^1x^6sqrt(1-x^2)`dx

Answer

To evaluate the integral \( \int_0^1 x^6 \sqrt{1 - x^2} \, dx \), we will use a trigonometric substitution. Here are the steps to solve the integral:

### Step 1: Trigonometric Substitution
Let \( x = \sin \theta \). Then, the differential \( dx \) is given by:
\[ dx = \cos \theta \, d\theta \]
Also, we need to change the limits of integration:
- When \( x = 0 \), \( \theta = 0 \)
- When \( x = 1 \), \( \theta = \frac{\pi}{2} \)

### Step 2: Substitute in the Integral
Substituting \( x = \sin \theta \) into the integral, we have:
\[ \sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta \]
Thus, the integral becomes:
\[ \int_0^{\frac{\pi}{2}} (\sin \theta)^6 \cos \theta \cdot \cos \theta \, d\theta = \int_0^{\frac{\pi}{2}} \sin^6 \theta \cos^2 \theta \, d\theta \]

### Step 3: Use Wallis's Theorem
According to Wallis's theorem, for the integral of the form:
\[ \int_0^{\frac{\pi}{2}} \sin^m \theta \cos^n \theta \, d\theta \]
where \( m \) and \( n \) are non-negative integers, the result is given by:
\[ \frac{(m-1)(m-3)(m-5)\ldots}{(m+n)(m+n-2)(m+n-4)\ldots} \]

In our case, \( m = 6 \) and \( n = 2 \).

### Step 4: Calculate the Numerator and Denominator
Calculating the numerator:
\[ (m-1)(m-3) = 5 \cdot 3 \cdot 1 = 15 \]

Calculating the denominator:
\[ (m+n)(m+n-2) = 8 \cdot 6 = 48 \]

### Step 5: Final Calculation
Now, we can substitute these values into Wallis's formula:
\[ \int_0^{\frac{\pi}{2}} \sin^6 \theta \cos^2 \theta \, d\theta = \frac{15}{48} = \frac{5}{16} \]

### Step 6: Multiply by the Factor from the Cosine
Since we have \( \cos^2 \theta \) in the integral, we need to multiply the result by \( \frac{1}{2} \) (as per the adjustment for the cosine terms):
\[ \text{Final Result} = \frac{5}{16} \cdot \frac{1}{2} = \frac{5}{32} \]

Thus, the value of the integral \( \int_0^1 x^6 \sqrt{1 - x^2} \, dx \) is:
\[ \frac{5}{32} \]