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

MCQ

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

A
$\tan x + \cot x + c$
B
$\cot x - \tan x + c$
✓
$\tan x - \cot x + c$
D
None of these

✓

Answer

Correct option: C.

$\tan x - \cot x + c$

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

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