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  • Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 2 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 2 Maths Textbook Solution.

Answers (1)

Answer:  The required value of the integral is,

I=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\cot \theta-1}{\sqrt{2} \cot \theta}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{\cot \theta+1-\sqrt{2} \cot \theta}{\cot \theta+1+\sqrt{2} \cot \theta}\right|+c

 

Hint:  Use the identity formula \int \frac{1}{x^{2}-a^{2}}dx=\frac{1}{2a}\log\left ( \frac{x-a}{x+a} \right )+c

 

Given:I=\int \sqrt{\cot \theta d \theta}

Solution: Suppose \cot \theta \; asx^{2}

Differentiating the above equation with respect to θ.

\begin{aligned} -\operatorname{cosec}^{2} \theta d \theta &=2 x d x \\ d \theta &=-\frac{2 x}{1+\cot ^{2} \theta} d \theta \\ d \theta &=-\frac{2 x}{1+x^{4}} d x \\ I &=\int-\frac{2 x}{1+x^{4}} d x \quad\quad\quad\quad\quad \ldots .1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta \\ \end{aligned}

 Re-writing the given equation as,

I=\int \frac{1+\frac{1}{x^{2}}+1-\frac{1}{x^{2}}}{\frac{1}{x^{2}}+x^{2}} d x-\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+2} d x-\int \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-2}     [Making the perfect square as (a+b)^{2} ]

\begin{aligned} &\text { Let, } x-\frac{1}{x}=t \text { and } x+\frac{1}{x}=z \text { so, } \\ &I=-\int \frac{d t}{t^{2}+2}-\int \frac{d z}{z^{2}-2} \end{aligned}

 

On using identity,

 

=-\frac{1}{2}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c

The required value of the integral is,

=-\frac{1}{2}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c

Now, substitutex=\sqrt{}\cot \thetainto the above equation we get,

I=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\cot \theta-1}{\sqrt{2} \cot \theta}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{\cot \theta+1-\sqrt{2} \cot \theta}{\cot \theta+1+\sqrt{2} \cot \theta}\right|+c

Where c is the integrating constant.

Posted by

infoexpert27

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