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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 56 Maths Textbook Solution.

Answers (1)

Answer:

$\frac{1}{5} \log \left|\frac{\tan x-2}{2 \tan x+1}\right|+c$

Given:

$\int \frac{1}{(\sin x-2 \cos x)(2 \sin x+\cos x)} d x$

Hint:

To solve this statement we have to evaluate the team by partial fraction.

Solution:

$\int \frac{d x}{(\sin x-2 \cos x)(2 \sin x+\cos x)}$

Dividing cos²x with numerator and denominator.

$I=\frac{\frac{1}{\cos ^{2} x}}{\frac{\sin x-2 \cos x}{\cos x} \times \frac{2 \sin x+\cos x}{\cos x}}$

$I=\int \frac{\sec ^{2} x}{(\tan x-2)(2 \tan x+1)} d x$

$I=\int \frac{d t}{(t-2)(2 t+1)}$ $\left[\because t=\tan x, d t=\sec ^{2} x d x\right]$

$\frac{1}{(t-1)(2 t+1)}=\frac{a}{t-2}+\frac{b}{2 t+1}$

$=\frac{(2 a+b) t+(a-2 b)}{(t-2)(2 t+1)}$

$2 a+b=0 \ldots \ldots \ldots \ldots \text { (1) }$

$a-2 b=1 \ldots \ldots \ldots \ldots(2)$

$\text { --------------------}$

$1 \times 2+(2)=$

$4 a+a=1, \quad a=\frac{1}{5}$

$b=-\frac{2}{5}$

$I=\int\left[\frac{1}{5} \times \frac{1}{t-2}-\frac{2}{5} \times \frac{1}{2 t+1}\right] d t$

$I=\frac{1}{5} \log |t-2|-\frac{2}{5} \times \frac{1}{2} \log (2 t+1)+c$

$I=\frac{1}{5} \log \left|\frac{t-2}{2 t+1}\right|+c$

$I=\frac{1}{5} \log \left|\frac{\tan x-2}{2 \tan x+1}\right|+c$

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