# ∫ 1 Cos 2 X ( 1 − Tan X ) 2 D X - Mathematics

Sum
$\int\frac{1}{\text{cos}^2\text{ x }\left( 1 - \text{tan x} \right)^2} dx$

#### Solution

$\text{Let I } = \int\frac{1}{\cos^2 x \left( 1 - \tan x \right)^2}dx$
$= \int\frac{\sec^2 x}{\left( 1 - \tan x \right)^2} \text{dx}$
$= \int\frac{\sec^2 \text{x dx}}{\left( 1 - \tan x \right)^2}$

Let 1-  tan x = t

$- \sec^2 \text{x dx} = dt$
$\Rightarrow \sec^2\text{ x dx} = - dt$
$\therefore I = \int\frac{- dt}{t^2}$
$= - \int t^{- 2} dt$
$= - \left[ \frac{t^{- 2 + 1}}{- 2 + 1} \right] + C$
$= \frac{1}{t} + C$
$= \frac{1}{1 - \tan x} + C$
Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.3 | Q 19 | Page 24

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