Let the integral `I=int((2020)^(x+sin^(-1)(2020)^(x)))/(sqrt(1-(2020)^(2x)))dx =K^(2)(202)^(sin^(-1)(2020)^(x))+lambda` (where, `lambda` is constant of integration), then the value of `2020^(K)` is

Options

A. 2020
B. 2019
C. e
D. `(1)/(e )`

Answer

To solve the integral \( I = \int \frac{2020^{x + \sin^{-1}(2020^x)}}{\sqrt{1 - (2020^{2x})}} \, dx \) and find the value of \( 2020^K \), we will follow these steps:

### Step 1: Substitute \( t = 2020^x \)
Let \( t = 2020^x \). Then, the differential \( dx \) can be expressed as:
\[ dx = \frac{dt}{t \ln(2020)} \]

### Step 2: Rewrite the integral
Substituting \( t \) into the integral, we have:
\[ I = \int \frac{t \cdot 2020^{\sin^{-1}(t)}}{\sqrt{1 - t^2}} \cdot \frac{dt}{t \ln(2020)} = \frac{1}{\ln(2020)} \int \frac{2020^{\sin^{-1}(t)}}{\sqrt{1 - t^2}} \, dt \]

### Step 3: Substitute \( u = \sin^{-1}(t) \)
Now, let \( u = \sin^{-1}(t) \), which implies \( t = \sin(u) \). The differential \( dt \) is:
\[ dt = \cos(u) \, du \]
Thus, we can rewrite the integral as:
\[ I = \frac{1}{\ln(2020)} \int \frac{2020^u}{\sqrt{1 - \sin^2(u)}} \cos(u) \, du \]
Since \( \sqrt{1 - \sin^2(u)} = \cos(u) \), the integral simplifies to:
\[ I = \frac{1}{\ln(2020)} \int 2020^u \, du \]

### Step 4: Integrate
The integral of \( 2020^u \) is:
\[ \int 2020^u \, du = \frac{2020^u}{\ln(2020)} + C \]
Thus, we have:
\[ I = \frac{1}{\ln(2020)} \left( \frac{2020^{\sin^{-1}(t)}}{\ln(2020)} + C \right) \]
Substituting back \( t = 2020^x \):
\[ I = \frac{2020^{\sin^{-1}(2020^x)}}{\ln(2020)^2} + C \]

### Step 5: Compare with the given expression
We are given that:
\[ I = K^2 \cdot 2020^{\sin^{-1}(2020^x)} + \lambda \]
By comparing both expressions, we can see that:
\[ K^2 = \frac{1}{\ln(2020)^2} \]
Thus, taking the square root:
\[ K = \frac{1}{\ln(2020)} \]

### Step 6: Find \( 2020^K \)
Now we need to find \( 2020^K \):
\[ 2020^K = 2020^{\frac{1}{\ln(2020)}} \]
Using the property of logarithms, we can express this as:
\[ 2020^K = e^{\ln(2020) \cdot \frac{1}{\ln(2020)}} = e \]

### Final Answer
Thus, the value of \( 2020^K \) is:
\[ \boxed{e} \]