Question

The indefinite integral $I=\displaystyle \int \frac{\left(\left(sin\right)^{2} x - \left(cos\right)^{2} ⁡ x\right)^{2019}}{\left(sin ⁡ x\right)^{2021} \left(cos ⁡ x\right)^{2021}}dx$ 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. $\frac{\left(tan x - cot ⁡ x\right)^{2020}}{2020}+c$
• 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)

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