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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 12

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 12

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Answer : -       -\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{1-x^{2}}{2 x^{2}}}\right)+C

Hint :-                Use substitution method and special integration formula..

Given :-                    \int \frac{1}{\left(x^{2}+1\right) \sqrt{1-x^{2}}} d x

Sol : -             Let  \mathrm{I}=\int \frac{1}{\left(x^{2}+1\right) \sqrt{1-x^{2}}} d x          

                                 Put  \mathrm{x}=\frac{1}{t} \rightarrow \mathrm{d} \mathrm{x}=-\frac{1}{t^{2}} \mathrm{dt} then,

\begin{aligned} I &=\int \frac{1}{\left(1+\frac{1}{t^{2}}\right) \sqrt{1-\frac{1}{t^{2}}}}\left(-\frac{1}{t^{2}}\right) d t \\ &=-\int \frac{\frac{1}{t^{2}}}{\left(\frac{t^{2}+1}{t^{2}}\right) \sqrt{\frac{t^{2}-1}{t^{2}}}} d t \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}}\left(t^{2}+1\right) \cdot \frac{1}{t} \sqrt{t^{2}-1}} d t \end{aligned}

                                                 =-\int \frac{t}{\left(t^{2}+1\right) \sqrt{t^{2}-1}} \mathrm{dt}

   Put,  t^{2}-1=u^{2} \rightarrow 2 \mathrm{t} \mathrm{dt}=2 \mathrm{u} \mathrm{du} \rightarrow \mathrm{t} \mathrm{dt}=\mathrm{u} \mathrm{du} than,

 \begin{aligned} &\quad I=-\int \frac{u d u}{\left(u^{2}+1+1\right) \sqrt{u^{2}}} \quad\left[\because t^{2}-1=u^{2} \rightarrow t^{2}=u^{2}+1\right] \\ &=-\int \frac{u d u}{\left(u^{2}+2\right) u} \\ &\quad=-\int \frac{d u}{u^{2}+(\sqrt{2})^{2}} \end{aligned}

    \begin{aligned} &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+\mathrm{c} \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+\mathrm{c}\right] \\ &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\sqrt{t^{2}-1}}{\sqrt{2}}\right)+\mathrm{c} \quad\left[\because u^{2}=t^{2}-1 \rightarrow \mathrm{u}=\sqrt{t^{2}-1}\right] \\ &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{t^{2}-1}{2}}\right)+\mathrm{c} \end{aligned}

\begin{aligned} &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{\frac{1}{x^{2}}-1}{2}}\right)+\mathrm{c} \quad\left[\because \mathrm{x}=\frac{1}{t} \rightarrow \mathrm{t}=\frac{1}{x}\right]\\ &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{\frac{1-x^{2}}{x^{2}}}{2}}\right)+\mathrm{c}\\ &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{1-x^{2}}{2 x^{2}}}\right)+\mathrm{c} \end{aligned}

    

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