Question

Let the integral $I=\displaystyle \int \frac{\left(2020\right)^{x + \left(sin\right)^{- 1} \left(2020\right)^{x}}}{\sqrt{1 - \left(2020\right)^{2 x}}}dx$ $=K^{2}\left(2020\right)^{\left(sin\right)^{- 1} \left(2020\right)^{x}}+\lambda $ (where, $\lambda $ is the constant of integration), then the vaue of $2020^{K}$ is

• A. $2020$
• B. $2019$
• C. e
• D. $\frac{1}{e}$

Answer 1

Answer: (C)

Let, $\left(sin\right)^{- 1} \left(\left(2020\right)^{x}\right) = t$
$\Rightarrow \frac{\left(2020\right)^{x} ln \left(2020\right)}{\sqrt{1 - \left(2020\right)^{2 x}}}dx=dt$
$\Rightarrow I=\displaystyle \int \frac{\left(2020\right)^{t}}{ln \left(2020\right)} d t$
$=\frac{\left(2020\right)^{t}}{l n \left(2020\right)^{2}}+\lambda $
$=\frac{\left(2020\right)^{\left(sin\right)^{- 1} \left(2020\right)^{x}}}{\left(l n 2020\right)^{2}}+\lambda $
$\therefore K=\frac{1}{l n 2020}=log_{2020}e$
$\therefore 2020^{K}=2020^{log_{2020} e}=e$