Question

$\int \frac{{\text{cosec}}^{2}x-2017}{{\cos }^{2017}x}dx$ is equal to

• A. $-\frac{\text{cosec}x}{(\cos x{)}^{2016}}+C$
• B. $\frac{\cot x}{(\cos x{)}^{2017}}+C$
• C. $-\frac{\cot x}{(\cos x{)}^{2016}}$
• D. $\frac{\tan x}{(\cos x{)}^{2017}}+C$

Answer 1

Answer: (A)

$I=\int (\cos \stackrel{I}{x}{)}^{-2017}\cdot \stackrel{II}{cose{c}^{2}x}dx-2017\int (\cos x{)}^{-2017}dx$
$=(-\cot x)\cdot (\cos x{)}^{-2017}-\int (-2017)\cdot (\cos x{)}^{-2018}\cdot (-\sin x)(-\cot x)dx-2017\int (\cos x{)}^{-2017}dx$
$I=\frac{-\cot x}{(\cos x{)}^{2017}}+2017\int (\cos x{)}^{-2017}dx-\int (2017)(\cos x{)}^{-2017}dx$
$I=-\frac{\mathrm{cosec}x}{(\cos x{)}^{2016}}+C$