Question

$\int \frac{{cosec}^{2}x-2005}{{\cos }^{2005}x}dx$ is equal to

• A. $\frac{\cot x}{(\cos x{)}^{2005}}+c$
• B. $\frac{\tan x}{(\cos x{)}^{2005}}+c$
• C. $\frac{(-\tan x)}{(\cos x{)}^{2005}}+c$
• D. none of these

Answer 1

Answer: (D)

none of these

$\int \frac{{\csc }^{2}x-2005}{{\cos }^{2005}x}dx$

$=\int \left({\cos }^{-2005}x\right){\csc }^{2}xdx-2005\int \frac{dx}{{\cos }^{2005}x}$

$=\left(\cos x{)}^{-2005}\right(-\cot x)-\int (-2005\left)\right(\cos x{)}^{-2006}(-\sin x)(-\cot x)dx-2005\int \frac{dx}{{\cos }^{2005}x}$ ...... [Using integration by parts]

$=-\frac{\cot x}{(\cos x{)}^{2005}}+C$