Question

The indefinite integral $I=\int \frac{\left((\sin )^{2} x-(\cos )^{2} x\right)^{2019}}{(\sin x)^{2021}(\cos x)^{2021}} d x$ 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)

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