`int_(0)^(1)(x^(7))/(sqrt(1-x^(4)))dx` is equal to

Options

A. `1`
B. `1/3`
C. `2/3`
D. `(pi)/3`

Explanation

To solve the integral

\[ I = \int_{0}^{1} \frac{x^7}{\sqrt{1 - x^4}} \, dx, \]

we will use the substitution \( x^2 = \sin \theta \).

### Step 1: Substitute \( x^2 = \sin \theta \)

From the substitution, we have:

\[ x = \sqrt{\sin \theta} \quad \text{and} \quad dx = \frac{1}{2\sqrt{\sin \theta}} \cos \theta \, d\theta. \]

### Step 2: Change the limits of integration

When \( x = 0 \):

\[ \sin \theta = 0 \implies \theta = 0. \]

When \( x = 1 \):

\[ \sin \theta = 1 \implies \theta = \frac{\pi}{2}. \]

Thus, the limits change from \( 0 \) to \( \frac{\pi}{2} \).

### Step 3: Substitute in the integral

Now substituting \( x^2 = \sin \theta \) into the integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{(\sin \theta)^{7/2}}{\sqrt{1 - (\sin \theta)^2}} \cdot \frac{1}{2\sqrt{\sin \theta}} \cos \theta \, d\theta. \]

### Step 4: Simplify the integrand

We know that \( \sqrt{1 - \sin^2 \theta} = \cos \theta \). Therefore, we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{7/2} \theta}{\cos \theta} \cdot \frac{1}{2\sqrt{\sin \theta}} \cos \theta \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin^{6} \theta \, d\theta. \]

### Step 5: Evaluate the integral

The integral \( \int_{0}^{\frac{\pi}{2}} \sin^{6} \theta \, d\theta \) can be evaluated using the Beta function or the formula:

\[ \int_{0}^{\frac{\pi}{2}} \sin^{n} \theta \, d\theta = \frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2} + 1\right)}. \]

For \( n = 6 \):

\[ \int_{0}^{\frac{\pi}{2}} \sin^{6} \theta \, d\theta = \frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma\left(\frac{7}{2}\right)}{\Gamma(4)}. \]

### Step 6: Calculate \( \Gamma \) values

We know:

\[ \Gamma(4) = 3! = 6, \]
\[ \Gamma\left(\frac{7}{2}\right) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15}{8} \sqrt{\pi}. \]

### Step 7: Substitute back

Thus,

\[ \int_{0}^{\frac{\pi}{2}} \sin^{6} \theta \, d\theta = \frac{\sqrt{\pi}}{2} \cdot \frac{\frac{15}{8} \sqrt{\pi}}{6} = \frac{15 \pi}{96}. \]

### Step 8: Final calculation

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

\[ I = \frac{1}{2} \cdot \frac{15 \pi}{96} = \frac{15 \pi}{192}. \]

Thus, the value of the integral is

\[ \boxed{\frac{15 \pi}{192}}. \]

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