Question

The indefinite integral $I=\int \frac{{\left({\left(sin\right)}^{2}x-{\left(cos\right)}^{2}x\right)}^{2019}}{{\left(sinx\right)}^{2021}{\left(cosx\right)}^{2021}}dx$ 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. $\frac{{\left(tanx-cotx\right)}^{2020}}{2020}+c$
• C. $\frac{{\left(sinx-cosx\right)}^{2020}}{2020}+c$
• D. $\frac{{\left({\left(tan\right)}^{2}x+{\left(cot\right)}^{2}x\right)}^{2020}}{2020}+c$

Answer 1

Answer: (B)

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