If `intsqrt((x)/(a^(3)-x^(3)))dx=msin^(-1)((x)/(a))^(n)+C`, then

Options

A. m = n
B. m =-n
C. `m=1//n`
D. `m=-1//n`

Explanation

To solve the integral \( \int \sqrt{\frac{x}{a^3 - x^3}} \, dx \) and relate it to the expression \( m \sin^{-1}\left(\frac{x}{a}\right)^n + C \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \sqrt{\frac{x}{a^3 - x^3}} \, dx \]
This can be rewritten as:
\[ \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} \, dx \]

### Step 2: Substitution
Now, we will use the substitution \( x^{3/2} = t \). Differentiating gives:
\[ \frac{3}{2} x^{1/2} \, dx = dt \quad \Rightarrow \quad dx = \frac{2}{3} t^{-1/3} \, dt \]
Substituting \( x = t^{2/3} \) into the integral, we have:
\[ \int \frac{\sqrt{t^{2/3}}}{\sqrt{a^3 - t^2}} \cdot \frac{2}{3} t^{-1/3} \, dt \]
This simplifies to:
\[ \int \frac{2}{3} \frac{t^{1/3}}{\sqrt{a^3 - t^2}} \, dt \]

### Step 3: Recognize the Integral Form
The integral now resembles the form:
\[ \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C \]
In our case, we can identify:
\[ \int \frac{t^{1/3}}{\sqrt{a^3 - t^2}} \, dt \]
This integral can be evaluated using the standard form.

### Step 4: Solve the Integral
Using the known integral form, we find:
\[ \int \frac{t^{1/3}}{\sqrt{a^3 - t^2}} \, dt = \sin^{-1}\left(\frac{t}{a^{3/2}}\right) + C \]
Thus, substituting back \( t = x^{3/2} \):
\[ \int \sqrt{\frac{x}{a^3 - x^3}} \, dx = \frac{2}{3} \sin^{-1}\left(\frac{x^{3/2}}{a^{3/2}}\right) + C \]

### Step 5: Identify m and n
From the expression, we can compare it with \( m \sin^{-1}\left(\frac{x}{a}\right)^n + C \):
- We have \( m = \frac{2}{3} \)
- The argument of the sine inverse function suggests \( n = \frac{3}{2} \)

### Conclusion
Thus, we find the relationship:
\[ m = \frac{2}{3}, \quad n = \frac{3}{2} \]
This gives us the relation:
\[ m = \frac{2}{3}, \quad n = \frac{3}{2} \quad \Rightarrow \quad m = \frac{1}{n} \]