MCQ
$\int_{}^{} {{{(\tan x - \cot x)}^2}\;dx = } $
- A$\tan x + \cot x + c$
- B$\sec x\tan x + c$
- C$\cos {\rm{ec}}x\cot x + c$
- ✓None of these
✓
Answer
Correct option: D.
None of these
d
(d)$\int_{}^{} {{{(\tan x - \cot x)}^2}dx} = \int_{}^{} {({{\tan }^2}x + {{\cot }^2}x - 2)\,dx} $
$ = \int_{}^{} {{{\sec }^2}x\,dx} + \int_{}^{} {{\rm{cose}}{{\rm{c}}^{\rm{2}}}x\,dx} - \int_{}^{} {4\,dx} $
$ = \tan x - \cot x - 4x + c.$
(d)$\int_{}^{} {{{(\tan x - \cot x)}^2}dx} = \int_{}^{} {({{\tan }^2}x + {{\cot }^2}x - 2)\,dx} $
$ = \int_{}^{} {{{\sec }^2}x\,dx} + \int_{}^{} {{\rm{cose}}{{\rm{c}}^{\rm{2}}}x\,dx} - \int_{}^{} {4\,dx} $
$ = \tan x - \cot x - 4x + c.$
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