#### Please solve RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 3 maths textbook solution.

$\\I=-\frac{2}{3}\left(2+3 x-x^{2}\right)^{3}-\left(\frac{2 x-3}{2}\right) \sqrt{2+3 x-x^{2}}-\frac{17}{4} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{17}}\right)+C$

Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant

Given : $\int(2 x-5) \sqrt{2+3 x-x^{2}} d x$

Solution :

\begin{aligned} &2 x-5=-(5-2 x)=-(2+3-2 x)=-2-(3-2 x) \\ &I=\int(-2-(3-2 x)) \sqrt{2+3 x-x^{2}} d x \\ &I=\int-2 \sqrt{2+3 x-x^{2}} d x-\int(3-2 x) \sqrt{2+3 x-x^{2}} d x \end{aligned}

Again, $2+3 x-x^{2}$

\begin{aligned} &=-\left(x^{2}-3 x-2\right) \\ &=-\left(x^{2}-2 \cdot x \cdot \frac{3}{2}+\frac{9}{4}-2-\frac{9}{4}\right) \\ &=-\left(\left(x-\frac{3}{2}\right)^{2}-\frac{17}{4}\right) \\ &=\left(\left(\sqrt{\frac{17}{4}}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}\right) \end{aligned}

Now, $I=-2 \int\left[\sqrt{\left(\sqrt{\frac{17}{4}}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}\right] \mathrm{dx}-\int(3-2 x) \sqrt{2+3 x-x^{2}} d x$

For the second integral :

Let, $2+3 x-x^{2}=t^{2}$

$\Rightarrow[(3-2 x) d x=2 t d t]$

Use the formula : $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$

\begin{aligned} &I=-2\left[\frac{\left(x-\frac{3}{2}\right)}{2} \sqrt{\frac{17}{4}-\left(x-\frac{3}{2}\right)^{2}}+\frac{\frac{17}{4}}{2} \sin ^{-1} \frac{\left(x-\frac{3}{2}\right)}{\frac{17}{4}}\right]-\int 2 t^{2} d t \\ &I=-\left(\frac{2 x-3}{2}\right) \sqrt{2+3 x-x^{2}}-\frac{17}{4} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{17}}\right)-\frac{2}{3}\left(2+3 x-x^{2}\right)^{3}+C \\ &I=-\frac{2}{3}\left(2+3 x-x^{2}\right)^{3}-\left(\frac{2 x-3}{2}\right) \sqrt{2+3 x-x^{2}}-\frac{17}{4} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{17}}\right)+C \end{aligned}