_ ^ 1 + ^2 x ^2 x dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}dx} = $

A
$ - \cot x - 2x + c$
B
$ - 2\cot x - 2x + c$
✓
$ - 2\cot x - x + c$
D
$ - 2\cot x + x + c$

✓

Answer

Correct option: C.

$ - 2\cot x - x + c$

c
(c) $\int_{}^{} {\frac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}} \,dx = \int_{}^{} {({\rm{cose}}{{\rm{c}}^2}x + {{\cot }^2}x)\,dx} $
$ = \int_{}^{} {(2{\rm{cose}}{{\rm{c}}^2}x - 1)\,dx = - 2\cot x - x + c.} $

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