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

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}

$=a(\sqrt{2}-1)$