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  • Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 4 maths

Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 4 maths

Answers (1)

Answer: a(\sqrt{2}-1)

Hint: Using \int x^{n} d x

Given:\int_{0}^{a} \frac{x}{\sqrt{a^{2}+x^{2}}} d x

Solution:  

I=\int_{0}^{a} \frac{x}{\sqrt{a^{2}+x^{2}}} d x

Put

    \begin{aligned} &a^{2}+x^{2}=t \\\\ &2 x d x=d t \\\\ &x d x=\frac{d t}{2} \end{aligned}

When x = 0, then t = a^{2}

When x = a, then t = 2a^{2}

I=\int_{a^{2}}^{2 a^{2}} \frac{1}{\sqrt{t}} \cdot \frac{d t}{2}

    \begin{aligned} &=\frac{1}{2} \int_{a^{2}}^{2 a^{2}} \frac{1}{\sqrt{t}} d t \\\\ &=\frac{1}{2} \int_{a^{2}}^{2 a^{2}}(t)^{\frac{-1}{2}} d t \end{aligned}

    \begin{aligned} &=\frac{1}{2}\left[\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\right]_{a^{2}}^{2 a^{2}} \\\\ &=\frac{1}{2}\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right]_{a^{2}}^{2 a^{2}} \end{aligned}

    \begin{aligned} &=\left[t^{\frac{1}{2}}\right]_{a^{2}}^{2 a^{2}} \\\\ &=\left(2 a^{2}\right)^{\frac{1}{2}}-\left(a^{2}\right)^{\frac{1}{2}} \\\\ &=(\sqrt{2} a)-(a) \end{aligned}

    =a(\sqrt{2}-1)

 

 

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