For `a gt 0, if I = int sqrt((x)/(a ^(3) -x ^(3)))dx = A sin ^(-1) ((x ^(3//2))/(B))+ C,` where C is any arbitary constant, then:

Options

A. `A=2/3`
B. `B=a ^(3//2)`
C. `A=1/3`
D. `B =a ^(1//2)`

Answer

To solve the integral \( I = \int \sqrt{\frac{x}{a^3 - x^3}} \, dx \) and express it in the form \( A \sin^{-1}\left(\frac{x^{3/2}}{B}\right) + C \), where \( C \) is an arbitrary constant, we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} \, dx \]

### Step 2: Substitution
Let \( t = x^{3/2} \). Then, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = \frac{3}{2} x^{1/2} \quad \Rightarrow \quad dx = \frac{2}{3} x^{-1/2} dt \]
Since \( x = t^{2/3} \), we have:
\[ dx = \frac{2}{3} t^{-1/3} dt \]

### Step 3: Substitute in the Integral
Substituting \( x \) and \( dx \) into the integral, we get:
\[ I = \int \frac{\sqrt{t^{2/3}}}{\sqrt{a^3 - (t^{2/3})^3}} \cdot \frac{2}{3} t^{-1/3} dt \]
This simplifies to:
\[ I = \int \frac{t^{1/3}}{\sqrt{a^3 - t^2}} \cdot \frac{2}{3} t^{-1/3} dt = \frac{2}{3} \int \frac{1}{\sqrt{a^3 - t^2}} dt \]

### Step 4: Recognize the Standard Integral
The integral \( \int \frac{1}{\sqrt{a^2 - t^2}} dt \) is a standard integral that evaluates to \( \sin^{-1}\left(\frac{t}{a}\right) + C \). Here, we have:
\[ I = \frac{2}{3} \sin^{-1}\left(\frac{t}{\sqrt{a^3}}\right) + C \]

### Step 5: Substitute Back for \( t \)
Recalling that \( t = x^{3/2} \), we substitute back:
\[ I = \frac{2}{3} \sin^{-1}\left(\frac{x^{3/2}}{\sqrt{a^3}}\right) + C \]

### Step 6: Identify Constants \( A \) and \( B \)
Comparing this with the given form \( A \sin^{-1}\left(\frac{x^{3/2}}{B}\right) + C \), we find:
- \( A = \frac{2}{3} \)
- \( B = \sqrt{a^3} \)

### Conclusion
Thus, the values of \( A \) and \( B \) are:
\[ A = \frac{2}{3}, \quad B = a^{3/2} \]