Get Answers to all your Questions

header-bg qa

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

Answers (1)

best_answer

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

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads