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  • need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 55

need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 55

Answers (1)

Answer: \frac{1}{6 a^{2}}\left[\left(x^{2}+a^{2}\right)^{\frac{3}{2}}-\left(x^{2}-a^{2}\right)^{\frac{3}{2}}\right]+c

Hint: Use substitution method to solve this integral.

Given:   \int \frac{x}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}-a^{2}}} d x

Solution:

        \text { Let } I=\int \frac{x}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}-a^{2}}} d x

        On Rationalising we get

        I=\int\left(\frac{x}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}-a^{2}}} \times \frac{\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}}{\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}}\right) d x

          =\int\left(\frac{x\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}{\left(\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}-a^{2}}\right)\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}\right) d x

        =\int \frac{x\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}{\left(\sqrt{x^{2}+a^{2}}\right)^{2}-\left(\sqrt{x^{2}-a^{2}}\right)^{2}} d x

        =\int \frac{x\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}{\left(x^{2}+a^{2}\right)-\left(x^{2}-a^{2}\right)} d x

        =\int \frac{x\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}{x^{2}+a^{2}-x^{2}+a^{2}} d x

        =\int \frac{x\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}{2 a^{2}} d x

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

        =\frac{1}{2 a^{2}} \int x \sqrt{x^{2}+a^{2}} d x-\frac{1}{2 a^{2}} \int x \sqrt{x^{2}-a^{2}} d x      ..........(i)

        \text { Now } \frac{1}{2 a^{2}} \int x \sqrt{x^{2}+a^{2}} d x

        \begin{aligned} &\text { Put } x^{2}+a^{2}=t \Rightarrow 2 x \; d x=d t \Rightarrow d x=\frac{a \imath}{2 x} \\ &\text { Then, } \frac{1}{2 a^{2}} \int x \sqrt{x^{2}+a^{2}} d x=\frac{1}{2 a^{2}} \int x \sqrt{t} \cdot \frac{d t}{2 x}=\frac{1}{4 a^{2}} \int t^{\frac{1}{2}} d t \end{aligned}

        =\frac{1}{4 a^{2}} \cdot \frac{t_{2}^{\frac{1}{1}+1}}{\frac{1}{2}+1}+c_{1}=\frac{1}{4 a^{2}} \cdot \frac{2}{3} \cdot t^{\frac{3}{2}}+c_{1} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

        =\frac{1}{6 a^{2}}\left(x^{2}+a^{2}\right)^{\frac{3}{2}}+c_{1}                                                        ......(ii)          \left(\because t=x^{2}+a^{2}\right)

        \text { and, } \frac{1}{2 a^{2}} \int x \sqrt{x^{2}-a^{2}} d x

        \begin{aligned} &\text { Put, } x^{2}-a^{2}=u \Rightarrow 2 x d x=d u \Rightarrow d x=\frac{d u}{2 x} \\ &\frac{1}{2 a^{2}} \int x \sqrt{x^{2}-a^{2}} d x=\frac{1}{2 a^{2}} \int x \cdot \sqrt{u} \cdot \frac{d u}{2 x}=\frac{1}{4 a^{2}} \int u_{}^{\frac{1}{2}} d u \text { then, } \end{aligned}

        =\frac{1}{4 a^{2}}\left[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+c_{2} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

        \begin{aligned} &=\frac{1}{4 a^{2}}\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]+c_{2} \\ &=\frac{1}{4 a^{2}} \cdot \frac{2}{3} u_{}^{\frac{3}{2}}+c_{2} \end{aligned}

        =\frac{1}{6 a^{2}}\left(x^{2}-a^{2}\right)^{\frac{3}{2}}+c_{2}                                                        ....(iii)          \left(\because u=x^{2}-a^{2}\right)

        Putting the values of eqn(ii) and eqn(iii) in (i) then,

        \begin{aligned} &I=\frac{1}{6 a^{2}}\left(x^{2}+a^{2}\right)^{\frac{3}{2}}+c_{1}-\frac{1}{6 a^{2}}\left(x^{2}-a^{2}\right)^{\frac{3}{2}}-c_{2} \\ &=\frac{1}{6 a^{2}}\left[\left(x^{2}+a^{2}\right)^{\frac{3}{2}}-\left(x^{2}-a^{2}\right)^{\frac{3}{2}}\right]+c \quad\left(\because c=c_{1}-c_{2}\right) \end{aligned}

 

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