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

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

#### Solution

$\text{ Let I } = \int \frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)}dx$
$\text{Dividing numerator and denominator by} \cos^2 x$
$\Rightarrow I = \int \frac{\sec^2 x}{\left( \frac{\sin x - 2 \cos x}{\cos x} \right) \times \left( \frac{2 \sin x + \cos x}{\cos x} \right)}dx$
$= \int \frac{\sec^2 x}{\left( \tan x - 2 \right) \left( 2 \tan x + 1 \right)}dx$
$\text{ Let tan x} = t$
$\Rightarrow \sec^2 x \text{ dx } = dt$
$\therefore I = \int \frac{dt}{\left( t - 2 \right) \left( 2t + 1 \right)}$

$= \int \frac{dt}{2 t^2 + t - 4t - 2}$
$= \int \frac{dt}{2 t^2 - 3t - 2}$
$= \frac{1}{2}\int \frac{dt}{t^2 - \frac{3}{2}t - 1}$
$= \frac{1}{2}\int \frac{dt}{t^2 - \frac{3}{2}t + \left( \frac{3}{4} \right)^2 - \left( \frac{3}{4} \right)^2 - 1}$
$= \frac{1}{2}\int \frac{dt}{\left( t - \frac{3}{4} \right)^2 - \frac{9}{16} - 1}$
$= \frac{1}{2}\int \frac{dt}{\left( t - \frac{3}{4} \right)^2 - \left( \frac{5}{4} \right)^2}$
$= \frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \text{ log } \left| \frac{t - \frac{3}{4} - \frac{5}{4}}{t - \frac{3}{4} + \frac{5}{4}} \right| + C$
$= \frac{1}{5} \text{ ln } \left| \frac{t - 2}{t + \frac{1}{2}} \right| + C$
$= \frac{1}{5}\text{ ln } \left| \frac{\left( t - 2 \right)^2}{2t + 1} \right| + C$
$= \frac{1}{5}\text{ln } \left| \frac{t - 2}{2t + 1} \right| + \frac{1}{5} \ln \left( 2 \right) + C$
$= \frac{1}{5} \text{ ln } \left| \frac{t - 2}{2t + 1} \right| + \text{ C where C} = C + \frac{1}{5}\ln \left( 2 \right)$
$= \frac{1}{5} \text{ ln } \left| \frac{\tan x - 2}{2 \tan x + 1} \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.22 | Q 7 | Page 114