The indefinite integral I = ∫ ( ^ 2 x - ^ 2 x) ^ 2019 (sin x) ^ 2021 (cos x) ^ 2021 dx simplifies to (where c is an integration constant)

Correct answer: B. (tan x - cot x) ^ 2020 2020 + c I = ∫ ( ^ 2 x - ^ 2 x sin xcos x ) ^ 2019 1 ^ 2 x ^ 2 x dx = ∫ (tan x - cot x) ^ 2019 ( ^ 2 x + ^ 2 x ^ 2 x ^ 2 x )dx = ∫ (tan x - cot x)2019( ^ 2 x + cosec^ 2 x)dx Let tan x - cot x = t ⇒ ( ^ 2 x + cosec ^ 2 x)dx = dt ∴ I = ∫ t^ 2019 dt = t^ 2020 2020 + c = (tan x - cot x) ^ 2020 2020 + c

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The indefinite integral I = ∫ ( ^ 2 x - ^ 2 x) ^ 2019 (sin x) ^...

The indefinite integral

I = ∫ \dfrac{{\left(\sin^{2} x - \cos^{2} x\right)}^{2019}}

{{

\left(sin x\right)}^{2021}{\left(cos x\right)}^{2021}}dx

simplifies to (where

c

is an integration constant)

Options

A $\dfrac{{\left(\sin^{2} x - \cos^{2} x\right)}^{2020}}{2020} + c$
B $\dfrac{{\left(tan x - cot x\right)}^{2020}}{2020} + c$
C $\dfrac{{\left(sin x - cos x\right)}^{2020}}{2020} + c$
D $\dfrac{{\left(\tan^{2} x + \cot^{2} x\right)}^{2020}}{2020} + c$

Correct answer

B. $\dfrac{{\left(tan x - cot x\right)}^{2020}}{2020} + c$

Step-by-step solution

I = {∫ \left(\dfrac{\sin^{2} x - \cos^{2} x}{sin xcos x}\right)}^{2019}\dfrac{1}{\sin^{2} x\cos^{2} x}dx

= ∫ {\left(tan x - cot x\right)}^{2019}\left(\dfrac{\sin^{2} x + \cos^{2} x}{\sin^{2} x\cos^{2} x}\right)dx

= ∫ \left(tan x - cot x\right)2019\left(\sec^{2} x + cosec^{2}x\right)dx

Let

tan x - cot x = t

⇒ \left(\sec^{2} x + {cosec}^{2}x\right)dx = dt

∴ I = ∫ t^{2019}dt

= \dfrac{t^{2020}}{2020} + c

= \dfrac{{\left(tan x - cot x\right)}^{2020}}{2020} + c

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