If `int( 2^(x))/( sqrt( 1- 4^(x))) dx = k sin^(-1) ( 2^(x)) + C`, then the value of k is

Options

A. `log 2`
B. ` (1)/( log 2 )`
C. `( 2)/( log 2)`
D. `( log 2)/( 2)`

Answer

To solve the integral \( \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx \) and find the value of \( k \) in the equation \( \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx = k \sin^{-1}(2^x) + C \), we will follow these steps:

### Step 1: Simplify the Integral
We start with the integral:
\[ \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx \]
Notice that \( 4^x = (2^2)^x = (2^x)^2 \). Therefore, we can rewrite the integral as:
\[ \int \frac{2^x}{\sqrt{1 - (2^x)^2}} \, dx \]

### Step 2: Substitution
Let \( u = 2^x \). Then, differentiating both sides gives:
\[ \frac{du}{dx} = 2^x \ln(2) \implies dx = \frac{du}{u \ln(2)} \]
Now, substituting \( u \) into the integral:
\[ \int \frac{u}{\sqrt{1 - u^2}} \cdot \frac{du}{u \ln(2)} = \frac{1}{\ln(2)} \int \frac{1}{\sqrt{1 - u^2}} \, du \]

### Step 3: Integrate
The integral \( \int \frac{1}{\sqrt{1 - u^2}} \, du \) is a standard integral that equals \( \sin^{-1}(u) + C \). Thus, we have:
\[ \int \frac{u}{\sqrt{1 - u^2}} \, du = \sin^{-1}(u) + C \]
Substituting back, we get:
\[ \frac{1}{\ln(2)} \sin^{-1}(u) + C = \frac{1}{\ln(2)} \sin^{-1}(2^x) + C \]

### Step 4: Compare with Given Expression
We know from the problem statement that:
\[ \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx = k \sin^{-1}(2^x) + C \]
By comparing both sides, we find:
\[ k = \frac{1}{\ln(2)} \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{\frac{1}{\ln(2)}} \]