Let the integral I = ∫ 2020 x + sin - 1 ⁡ 2020 x 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

Question

Let the integral I = ∫ 2020 x + sin - 1 ⁡ 2020 x 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, 2020, 2019, e, 1 e

MARKS App · Accepted Answer

Let, sin - 1 ⁡ 2020 x = t ⇒ 2020 x ln ⁡ 2020 1 - 2020 2 x d x = d t ⇒ I = ∫ 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 = log 2020 e ∴ 2020 K = 2020 log 2020 ⁡ e = e

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