Let the integral I=∫2020^(x+sin-1)⁡2020^x√(1-2020^2x)dx=K^22020^(sin-1)⁡2020^x+λ (where, λ is the constant of integration), then the vaue of 2020^K is

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) e. Let, sin^(-1)⁡2020^x=t ⇒(2020^xln⁡2020)/(√(1-2020^2x))dx=dt ⇒I=∫(2020^t)/(ln⁡2020)dt =2020^tln2020^2+λ =2020^(sin-1)⁡2020^xln2020^2+λ ∴K=1/ln2020=log_2020e ∴2020^K=2020^(log2020⁡e)=e

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

Solution

More Indefinite Integration Questions

Let the integral I=∫2020x+sin-1⁡2020x1-20202xdx=K22020sin-1⁡2020x+λ (where, λ is the constant of integration), then the vaue of 2020K is

Solution

More Indefinite Integration Questions

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