Question

The indefinite integral $I=\int \frac{{\left((\sin {)}^{2}x-(\cos {)}^{2}x\right)}^{2019}}{(\sin x{)}^{2021}(\cos x{)}^{2021}}dx$ simplifies to (where $c$ is an integration constant)

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

Answer 1

Answer: (B)

$I=\int \frac{{\sin }^{2}x-{\cos }^{2}x}{\sin x\cos x}\frac{1}{{\sin }^{2}x{\cos }^{2}x}dx$
$=\int \tan x-\cot x^{2019}\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$
$=\int \tan x-\cot x^{2019}{\sec }^{2}x+{\text{cosec}}^{2}xdx$
Let $\tan x-\cot x=t$
$\Rightarrow {\sec }^{2}x+{\text{cosec}}^{2}xdx=dt$
$\therefore I=\int t^{2019}dt$
$=\frac{t^{2020}}{2020}+c$
$=\frac{\tan x-\cot x^{2020}}{2020}+c$