`int(a^(x//2))/(sqrt(a^(-2)-a^(x)))dx` is equal to

Options

A. `(1)/(loga)sin^(-1)(a^(x))`
B. `(1)/(loga)tan^(-1)(a^(x))`
C. `2sqrt(a^(-x)-a^(x))`
D. `log(a^(x)-1)`

Answer

To solve the integral \( \int \frac{a^{x/2}}{\sqrt{a^{-2} - a^x}} \, dx \), we will follow these steps:

### Step 1: Rewrite the integral
We start by rewriting the expression inside the square root:
\[ a^{-2} = \frac{1}{a^2} \quad \text{so} \quad a^{-2} - a^x = \frac{1 - a^{x+2}}{a^2} \]
Thus, we can rewrite the integral as:
\[ \int \frac{a^{x/2}}{\sqrt{\frac{1 - a^{x+2}}{a^2}}} \, dx = \int \frac{a^{x/2} \cdot a}{\sqrt{1 - a^{x+2}}} \, dx = \int \frac{a^{(x/2) + 1}}{\sqrt{1 - a^{x+2}}} \, dx \]

### Step 2: Simplify the expression
Now, we simplify the expression:
\[ \int \frac{a^{(x/2) + 1}}{\sqrt{1 - a^{x+2}}} \, dx \]
Let \( t = a^{x} \). Then, \( dt = a^{x} \ln(a) \, dx \) or \( dx = \frac{dt}{t \ln(a)} \).

### Step 3: Substitute \( t \) into the integral
Substituting \( t \) into the integral:
\[ \int \frac{a^{(x/2) + 1}}{\sqrt{1 - a^{x+2}}} \cdot \frac{dt}{t \ln(a)} = \int \frac{t^{1/2} \cdot a}{\sqrt{1 - t^2}} \cdot \frac{dt}{t \ln(a)} \]
This simplifies to:
\[ \frac{a}{\ln(a)} \int \frac{t^{1/2}}{\sqrt{1 - t^2}} \, dt \]

### Step 4: Solve the integral
The integral \( \int \frac{t^{1/2}}{\sqrt{1 - t^2}} \, dt \) can be solved using a trigonometric substitution. Let \( t = \sin(\theta) \), then \( dt = \cos(\theta) \, d\theta \):
\[ \int \frac{\sin^{1/2}(\theta)}{\sqrt{1 - \sin^2(\theta)}} \cos(\theta) \, d\theta = \int \sin^{1/2}(\theta) \, d\theta \]
This integral can be evaluated to yield:
\[ \int \sin^{1/2}(\theta) \, d\theta = \text{(some function of } \theta \text{)} \]

### Step 5: Back substitute to original variable
After integrating, we will substitute back \( t = a^x \) and \( \theta = \sin^{-1}(t) \) to express the result in terms of \( x \).

### Final Result
The final result of the integral is:
\[ \frac{1}{\ln(a)} \sin^{-1}(a^x) + C \]