#### Please Solve RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.17 Question 5 Maths Textbook Solution

Answer $\sin^{-1}\left ( \frac{2x-\alpha-\beta}{\beta -\alpha } \right )+c$

Hint:-To solve this problem, use special integration formula

Given:- $\int \frac{1}{\sqrt{\left ( 1-\alpha \right )\left ( \beta -x \right )}}dx,\left ( \beta > \alpha \right )$

Solution:-

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

\begin{aligned} &=\int \frac{1}{\sqrt{-\left[\left\{x-\left(\frac{\alpha+\beta}{2}\right)^{2}\right\}-\left\{\frac{\alpha^{2}+\beta^{2}+2 \alpha \beta}{4}-\alpha \beta\right\}\right]}} d x \\ &=\int \frac{1}{\sqrt{-\left[\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}-\left\{\frac{\alpha^{2}+\beta^{2}+2 \alpha \beta-4 \alpha \beta}{4}\right\}\right]}} d x \\ &=\int \frac{1}{\sqrt{-\left[\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}-\left\{\frac{\alpha^{2}+\beta^{2}-2 \alpha \beta}{4}\right\}\right]}} d x \end{aligned}

\begin{aligned} &=\int \frac{1}{\sqrt{-\left\{\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}-\left(\frac{\alpha-\beta}{2}\right)^{2}\right]}} \Rightarrow \int \frac{1}{\sqrt{-\left[\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}-\left(\frac{\beta-\alpha}{2}\right)^{2}\right]} d x}\quad\quad\quad \quad(\because \beta>\alpha) \\ &=\int \frac{1}{\sqrt{\left(\frac{\beta-\alpha}{2}\right)^{2}-\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}}} d x \end{aligned}

\begin{aligned} &x-\left(\frac{\alpha+\beta}{2}\right)=t \Rightarrow d x=d t \text { then } \\ &\mathrm{I}=\int \frac{1}{\sqrt{\left(\frac{\beta-\alpha}{2}\right)^{2}-t^{2}}} d t=\sin ^{-1}\left(\frac{t}{\frac{\beta-\alpha}{2}}\right)+c \quad\quad\quad\quad\quad\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right] \end{aligned}

\begin{aligned} &=\sin ^{-1}\left(\frac{x-\left(\frac{\alpha+\beta}{2}\right)}{\frac{\beta-\alpha}{2}}\right)+c \Rightarrow \sin ^{-1}\left(\frac{2 x-(\alpha+\beta)}{\frac{2}{\frac{\beta-\alpha}{2}}}\right)+c\quad\quad\quad\quad\quad \quad\left[\because t=x-\left(\frac{\alpha+\beta}{2}\right)\right] \\ &=\sin ^{-1}\left(\frac{2 x-\alpha-\beta}{\beta-\alpha}\right)+c \end{aligned}