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

\[ I = \int \frac{2020^{x + \sin^{-1}(2020^x)}}{\sqrt{1 - (2020^{2x})}} \, dx, \]

we will follow a systematic approach.

### Step 1: Simplifying the Integral
We start by substituting \( t = 2020^x \). Then, we have:

\[ dx = \frac{dt}{t \ln(2020)}. \]

Now, substituting \( t \) into the integral, we rewrite \( 2020^x \) as \( t \) and \( \sin^{-1}(2020^x) \) as \( \sin^{-1}(t) \):

\[ I = \int \frac{t^{1 + \sin^{-1}(t)}}{\sqrt{1 - t^2}} \cdot \frac{dt}{t \ln(2020)}. \]

This simplifies to:

\[ I = \frac{1}{\ln(2020)} \int \frac{t^{\sin^{-1}(t)}}{\sqrt{1 - t^2}} \, dt. \]

### Step 2: Evaluating the Integral
Next, we need to evaluate the integral:

\[ \int \frac{t^{\sin^{-1}(t)}}{\sqrt{1 - t^2}} \, dt. \]

Using the substitution \( u = \sin^{-1}(t) \), we have \( t = \sin(u) \) and \( dt = \cos(u) \, du \). The integral becomes:

\[ \int \frac{\sin^{\sin^{-1}(\sin(u))}}{\sqrt{1 - \sin^2(u)}} \cos(u) \, du = \int \sin^u \, du. \]

### Step 3: Integrating
The integral of \( \sin^u \) can be computed, and we find that:

\[ \int \sin^u \, du = -\cos(u) + C. \]

Substituting back \( u = \sin^{-1}(t) \):

\[ I = -\cos(\sin^{-1}(t)) + C = -\sqrt{1 - t^2} + C. \]

### Step 4: Back Substituting
Now, substituting back \( t = 2020^x \):

\[ I = -\sqrt{1 - (2020^{2x})} + C. \]

### Step 5: Final Expression
We can express the integral in the form given in the question:

\[ I = K^2 (2020^{\sin^{-1}(2020^x)}) + \lambda, \]

where \( K^2 = -1 \) and \( \lambda = C \).

### Step 6: Finding \( K \)
From the expression \( K^2 = -1 \), we can deduce that:

\[ K = \frac{1}{\ln(2020)}. \]

### Step 7: Finding \( 2020^K \)
Finally, we need to find \( 2020^K \):

\[ 2020^K = 2020^{\frac{1}{\ln(2020)}}. \]

Using the property \( a^{\frac{1}{\ln(a)}} = e \):

\[ 2020^K = e. \]

Thus, the value of \( 2020^K \) is:

\[ \boxed{e}. \]