`"I f"intsqrt(x/(a^3-x^3))dx=d/bsin^(- 1)((x^(3/2))/(a^(3/2)))+C` (where `b & d` are coprime integer) then `b + d` equals to.

Options

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

Answer

To solve the integral \( \int \sqrt{\frac{x}{a^3 - x^3}} \, dx \) and express it in the form \( \frac{d}{b} \sin^{-1}\left(\frac{x^{3/2}}{a^{3/2}}\right) + C \), where \( b \) and \( d \) are coprime integers, we will follow these steps:

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

### Step 2: Substitution
Let \( t = x^{3/2} \). Then, we differentiate:
\[ x = t^{2/3} \implies dx = \frac{2}{3} t^{-1/3} dt \]
Substituting \( x \) and \( dx \) into the integral gives:
\[ I = \int \frac{\sqrt{t^{2/3}}}{\sqrt{a^3 - (t^{2/3})^3}} \cdot \frac{2}{3} t^{-1/3} dt \]
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 3: Integral of the Form
Now, we recognize that:
\[ \int \frac{1}{\sqrt{a^3 - t^2}} dt \]
is of the form \( \int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right) + C \).

### Step 4: Apply the Formula
Thus, we have:
\[ I = \frac{2}{3} \sin^{-1}\left(\frac{t}{\sqrt{a^3}}\right) + C \]
Substituting back \( t = x^{3/2} \):
\[ I = \frac{2}{3} \sin^{-1}\left(\frac{x^{3/2}}{a^{3/2}}\right) + C \]

### Step 5: Compare with the Given Form
According to the problem, we need to express this as:
\[ I = \frac{d}{b} \sin^{-1}\left(\frac{x^{3/2}}{a^{3/2}}\right) + C \]
From our result, we can see that \( d = 2 \) and \( b = 3 \).

### Step 6: Find \( b + d \)
Now, since \( b \) and \( d \) are coprime integers:
\[ b + d = 3 + 2 = 5 \]

### Final Answer
Thus, the value of \( b + d \) is:
\[ \boxed{5} \]