_ ^ 1 ^2 x (1 - x) ^2 dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{{{\cos }^2}x{{(1 - \tan x)}^2}}}dx = } $

A
$\frac{1}{{\tan x - 1}} + c$
✓
$\frac{1}{{1 - \tan x}} + c$
C
$ - \frac{1}{3}\frac{1}{{{{(1 - \tan x)}^3}}} + c$
D
None of these

✓

Answer

Correct option: B.

$\frac{1}{{1 - \tan x}} + c$

b
(b)$\int_{}^{} {\frac{1}{{{{\cos }^2}x{{(1 - \tan x)}^2}}}dx = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{(\tan x - 1)}^2}}}} } $
Put $\tan x - 1 = t \Rightarrow {\sec ^2}x\,dx = dt,$ then it reduces to
$\int_{}^{} {\frac{1}{{{t^2}}}\,dt} = \frac{{ - 1}}{{\tan x - 1}} + c = \frac{1}{{1 - \tan x}} + c.$

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