Evaluate the following integrals:
`int frac{sqrtx}{sqrt(a^3-x^3)}dx`

Answer

To evaluate the integral

\[ I = \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} \, dx, \]

we will use a substitution method to simplify the expression.

### Step 1: Substitution
Let us make the substitution \( x^{3/2} = t \). Then, we can express \( x \) in terms of \( t \):

\[ x = t^{2/3}. \]

Now, we need to find \( dx \) in terms of \( dt \). Differentiating both sides gives:

\[ dx = \frac{2}{3} t^{-1/3} \, dt. \]

### Step 2: Substitute in the Integral
Now we substitute \( x \) and \( dx \) into the integral:

\[ I = \int \frac{\sqrt{t^{2/3}}}{\sqrt{a^3 - (t^{2/3})^3}} \cdot \frac{2}{3} t^{-1/3} \, dt. \]

This simplifies to:

\[ I = \int \frac{t^{1/3}}{\sqrt{a^3 - t^2}} \cdot \frac{2}{3} t^{-1/3} \, dt. \]

### Step 3: Simplify the Integral
The \( t^{1/3} \) in the numerator and \( t^{-1/3} \) in the denominator cancel out:

\[ I = \frac{2}{3} \int \frac{1}{\sqrt{a^3 - t^2}} \, dt. \]

### Step 4: Recognize the Integral Form
The integral

\[ \int \frac{1}{\sqrt{a^3 - t^2}} \, dt \]

is a standard form that can be evaluated using the formula:

\[ \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + C. \]

In our case, we can set \( a = a^{3/2} \) and \( t \) corresponds to \( x \).

### Step 5: Evaluate the Integral
Thus, we have:

\[ I = \frac{2}{3} \cdot \sin^{-1}\left(\frac{t}{a^{3/2}}\right) + C. \]

### Step 6: Substitute Back to Original Variable
Recall that we made the substitution \( t = x^{3/2} \). Therefore, we substitute back to get:

\[ I = \frac{2}{3} \sin^{-1}\left(\frac{x^{3/2}}{a^{3/2}}\right) + C. \]

### Final Answer
We can rewrite the final answer in a more standard form:

\[ I = \frac{2}{3} \sin^{-1}\left(\frac{\sqrt{x^3}}{a^{3/2}}\right) + C. \]