cose c ^4 x ,dx = - Vidyadip

MCQ

$\int {{\rm{cose}}{{\rm{c}}^4}x\,dx} = $

A
$\cot x + \frac{{{{\cot }^3}x}}{3} + c$
B
$\tan x + \frac{{{{\tan }^3}x}}{3} + c$
✓
$ - \cot x - \frac{{{{\cot }^3}x}}{3} + c$
D
$ - \tan x - \frac{{{{\tan }^3}x}}{3} + c$

✓

Answer

Correct option: C.

$ - \cot x - \frac{{{{\cot }^3}x}}{3} + c$

c
(c) $ I = \int {cosec^4}x dx = \int {{\rm{cose}}{{\rm{c}}^2}x} .\,{\rm{cose}}{{\rm{c}}^2}xdx$
$ = \int {{\rm{cose}}{{\rm{c}}^2}x(1 + {{\cot }^2}x)\,dx} $
$ = \int {{\rm{cose}}{{\rm{c}}^2}x\,\,dx} \,\, + \int {{{\cot }^2}x.\,{\rm{cose}}{{\rm{c}}^2}x\,dx} $
$ = - \cot x - \frac{{{{\cot }^3}x}}{3} + c$.

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