Question

$\int {\cot }^{-1}(\sec x-\tan x)dx=\dots \dots$

• A. $\frac{\pi x}{4}-\frac{x^2}{4}+c$
• B. $\frac{\pi x}{4}+\frac{x^2}{4}+c$
• C. $\frac{x^2}{4}+c$
• D. $\frac{x^2}{2}+c$

Answer 1

Answer: (B)

$\frac{\pi x}{4}+\frac{x^2}{4}+c$

$I=\int {\cot }^{-1}(\sec x-\tan x)dx$

$=\int {\cot }^{-1}\left(\frac{1-\sin x}{\cos x}\right)$

$=\int {\cot }^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)dx$

$=\int {\cot }^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)dx$

$=\int {\cot }^{-1}\left(\tan (\frac{\pi }{4}-\frac{x}{2})\right)dx$

$=\int {\cot }^{-1}\left(\cot (\frac{\pi }{2}-\frac{\pi }{4}+\frac{x}{2})\right)dx$

$\Rightarrow I=\int (\frac{\pi }{4}+\frac{x}{2})dx$

$\Rightarrow I=\frac{\pi x}{4}+\frac{x^2}{4}+c$