`int(2^x)/(sqrt(1-4^x))dx=ksin^(- 1)2^x+c`, then k =

Options

A. log 2
B. `(1)/(2)log2`
C. `(1)/(2)`
D. `(1)/(log2)`

Answer

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

### Step 1: Rewrite the integral
We start by rewriting \( 4^x \) in terms of \( 2^x \):
\[ 4^x = (2^2)^x = (2^x)^2 \]
Thus, we can rewrite the integral as:
\[ \int \frac{2^x}{\sqrt{1 - (2^x)^2}} \, dx \]

### Step 2: Use substitution
Let \( t = 2^x \). Then, differentiating both sides gives:
\[ dt = 2^x \ln(2) \, dx \quad \Rightarrow \quad dx = \frac{dt}{2^x \ln(2)} = \frac{dt}{t \ln(2)} \]

### Step 3: Substitute in the integral
Substituting \( t \) and \( dx \) into the integral, we have:
\[ \int \frac{t}{\sqrt{1 - t^2}} \cdot \frac{dt}{t \ln(2)} = \frac{1}{\ln(2)} \int \frac{1}{\sqrt{1 - t^2}} \, dt \]

### Step 4: Integrate
The integral \( \int \frac{1}{\sqrt{1 - t^2}} \, dt \) is a standard integral that evaluates to \( \sin^{-1}(t) + C \). Therefore, we have:
\[ \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx = \frac{1}{\ln(2)} \sin^{-1}(t) + C \]

### Step 5: Substitute back for \( t \)
Substituting back \( t = 2^x \):
\[ \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx = \frac{1}{\ln(2)} \sin^{-1}(2^x) + C \]

### Step 6: Identify \( k \)
From the expression \( \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx = k \sin^{-1}(2^x) + c \), we can see that:
\[ k = \frac{1}{\ln(2)} \]

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