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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 62 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 62  Maths Textbook Solution.

Answers (1)

Answer:

\frac{1}{2 \sqrt{5}} \ln \left|\frac{\tan x-\left(\frac{1}{\sqrt{5}}\right)}{\tan x+\left(\frac{1}{\sqrt{5}}\right)}\right|+\mathrm{C}

Hint:

To solve the given statement we will take \cos 2 x \operatorname{as}\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)

Given:

\int \frac{1}{2-3 \cos 2 x} d x

Solution: 

I=\int \frac{1}{2-3\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)} d x

  I=\int \frac{1+\tan ^{2} x}{2\left(1+\tan ^{2} x\right)-3\left(1-\tan ^{2} x\right)} d x

I=\int \frac{\sec ^{2} x}{-1+\operatorname{stan}^{2} x} d x

I=\int \frac{\sec ^{2} x}{5\left(\tan ^{2} x-\frac{1}{5}\right)} d x

I=\int \frac{1}{5} \frac{d t}{t^{2}-\frac{1}{5}}                                        \left[\because \tan x=t, d t=\sec ^{2} x d x\right]

I=\frac{1}{5} \int \frac{d t}{t^{2}-\left(\frac{1}{\sqrt{5}}\right)^{2}}

I=\frac{1}{5} \frac{1}{2\left(\frac{1}{\sqrt{5}}\right)} \ln \left|\frac{t-\left(\frac{1}{\sqrt{5}}\right)}{t+\left(\frac{1}{\sqrt{5}}\right)}\right|+c

I=\frac{1}{2 \sqrt{5}} \ln \left|\frac{\tan x-\left(\frac{1}{\sqrt{5}}\right)}{\tan x+\left(\frac{1}{\sqrt{5}}\right)}\right|+\mathrm{C}

 

Posted by

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