#### Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 22

Hint: Using $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

Given: $\int_{-a}^{a} \sqrt{\frac{a-x}{a+x}} d x=K \pi$            ........(1)

Solution:

Let    $I=\int_{-a}^{a} \sqrt{\frac{a-x}{a+x}} d x$

$=\int_{-a}^{a} \sqrt{\frac{a-x}{a+x} \times \frac{a-x}{a-x}} d x=\int_{-a}^{a} \sqrt{\frac{(a-x)^{2}}{a^{2}-x^{2}}} d x$

$\therefore \mathrm{I}=\int_{-a}^{a} \frac{(a-x)}{\sqrt{a^{2}-x^{2}}} d x$            ........(2)

We have a property $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

Using this, we get:

\begin{aligned} \mathrm{I}=& \int_{-a}^{a} \frac{(a-(-x))}{\sqrt{a^{2}-\left(-x^{2}\right)}} d x \\\\ \Rightarrow \mathrm{I}=& \int_{-a}^{a} \frac{(a+x)}{\sqrt{a^{2}-x^{2}}} d x \end{aligned}            ............(3)

$2 \mathrm{I}=\int_{-a}^{a} \frac{(a-x)}{\sqrt{a^{2}-x^{2}}} d x \int_{+-a}^{a} \frac{(a+x)}{\sqrt{a^{2}-x^{2}}} d x$

$\begin{gathered} 2 \mathrm{I}=\int_{-a}^{a} \frac{a-x+a+x}{\sqrt{a^{2}-x^{2}}} d x \\\\\end{gathered}$

$2 \mathrm{I}=2 \int_{-a}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} d x$

$\mathrm{I}=\mathrm{a}\left[\sin ^{-1} \frac{x}{a}\right]_{-a}^{a} \quad\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}\right]$

\begin{aligned} &\mathrm{I}=\mathrm{a}\left[\sin ^{-1}\left(\frac{a}{a}\right)-\sin ^{-1}\left(\frac{-a}{a}\right)\right] \\\\ &\mathrm{I}=\mathrm{a}\left[\sin ^{-1}(1)-\sin ^{-1}(-1)\right] \end{aligned}

\begin{aligned} &\mathrm{I}=\mathrm{a}\left[\frac{\pi}{2}-\left(\frac{-\pi}{2}\right)\right] \\\\ &\mathrm{I}=\mathrm{a} \pi \quad \end{aligned}    .............(4)

On comparing (1) and (4), we get K = a.