Let the integral I= displaystyle ∫ ((2020)x + (sin)- 1 (2020)x/√1 - (2020)2 x)dx =K2(2020)(sin)- 1 (2020)x+λ (where, λ is the constant of integration), then the vaue of 2020K is

Question

Let the integral I= displaystyle ∫ ((2020)x + (sin)- 1 (2020)x/√1 - (2020)2 x)dx =K2(2020)(sin)- 1 (2020)x+λ (where, λ is the constant of integration), then the vaue of 2020K is (A) 2020 (B) 2019 (C) e (D) (1/e)

Tardigrade Admin · Accepted Answer

Let, (sin)- 1 ((2020)x) = t ⇒ ((2020)x ln (2020)/√1 - (2020)2 x)dx=dt ⇒ I= displaystyle ∫ ((2020)t/ln (2020)) d t =((2020)t/l n (2020)2)+λ =((2020)(sin)- 1 (2020)x/(l n 2020)2)+λ ∴ K=(1/l n 2020)=log2020e ∴ 2020K=2020log2020 e=e

Q. Let the integral $I = \int \frac{( 2020 ) ^{x + (s in)^{- 1} (2020)^{x}}}{1 - ( 2020 ) ^{2 x}} d x$ $= K^{2} (2020)^{(s in)^{- 1} (2020)^{x}} + λ$ (where, $λ$ is the constant of integration), then the vaue of $202 0^{K}$ is

$2020$

$2019$

e

$\frac{1}{e}$

Q. Let the integral I=∫1−(2020)2x​(2020)x+(sin)−1(2020)x​dx =K2(2020)(sin)−1(2020)x+λ (where, λ is the constant of integration), then the vaue of 2020K is

2020

2019

e

e1​

Let, (sin)−1((2020)x)=t ⇒1−(2020)2x​(2020)xln(2020)​dx=dt ⇒I=∫ln(2020)(2020)t​dt =ln(2020)2(2020)t​+λ =(ln2020)2(2020)(sin)−1(2020)x​+λ ∴K=ln20201​=log2020​e ∴2020K=2020log2020​e=e

Q. Let the integral $I = \int \frac{( 2020 ) ^{x + (s in)^{- 1} (2020)^{x}}}{1 - ( 2020 ) ^{2 x}} d x$ $= K^{2} (2020)^{(s in)^{- 1} (2020)^{x}} + λ$ (where, $λ$ is the constant of integration), then the vaue of $202 0^{K}$ is

$2020$

$2019$

$\frac{1}{e}$