Evaluate the following integrals -
$\int(2 x-5) \sqrt{2+3 x-x^{2}} d x$
Let $I=\int(2 x-5) \sqrt{2+3 x-x^{2}} d x$
Let us assume $2 x-5=\lambda \frac{d}{d x}\left(2+3 x-x^{2}\right)+\mu$
$\Rightarrow 2 x-5=\lambda\left[\frac{d}{d x}(2)+\frac{d}{d x}(3 x)-\frac{d}{d x}\left(x^{2}\right)\right]+\mu$
$\Rightarrow 2 x-5=\lambda\left[\frac{d}{d x}(2)+3 \frac{d}{d x}(x)-\frac{d}{d x}\left(x^{2}\right)\right]+\mu$
We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$ and derivative of a constant is 0 .
$\Rightarrow 2 x-5=\lambda\left(0+3-2 x^{2-1}\right)+\mu$
$\Rightarrow 2 x-5=\lambda(3-2 x)+\mu$
$\Rightarrow 2 x-5=-2 \lambda x+3 \lambda+\mu$
Comparing the coefficient of $x$ on both sides, we get
$-2 \lambda=2 \Rightarrow \lambda=-1$
Comparing the constant on both sides, we get
$3 \lambda+\mu=-5$
$\Rightarrow 3(-1)+\mu=-5$
$\Rightarrow-3+\mu=-5$
$\therefore \mu=-2$
Hence, we have $2 x-5=-(3-2 x)-2$
Substituting this value in I, we can write the integral as
$\mathrm{I}=\int[-(3-2 \mathrm{x})-2] \sqrt{2+3 \mathrm{x}-\mathrm{x}^{2}} \mathrm{dx}$
$\Rightarrow \mathrm{I}=\int\left[-(3-2 \mathrm{x}) \sqrt{2+3 \mathrm{x}-\mathrm{x}^{2}}-2 \sqrt{2+3 \mathrm{x}-\mathrm{x}^{2}}\right] \mathrm{dx}$
$\Rightarrow \mathrm{I}=-\int(3-2 \mathrm{x}) \sqrt{2+3 \mathrm{x}-\mathrm{x}^{2}} \mathrm{dx}-\int 2 \sqrt{2+3 \mathrm{x}-\mathrm{x}^{2}} \mathrm{dx}$
$\Rightarrow I=-\int(3-2 x) \sqrt{2+3 x-x^{2}} d x-2 \int \sqrt{2+3 x-x^{2}} d x$
Let $I_{1}=-\int(3-2 x) \sqrt{2+3 x-x^{2}} d x$
Now, put $2+3 x-x^{2}=t$Now, put $2+3 x-x^{2}=t$
$\Rightarrow(3-2 x) d x=d t$ (Differentiating both sides)
Substituting this value in $\mathrm{I}_{1}$, we can write
$\mathrm{I}_{1}=-\int \sqrt{\mathrm{t}} \mathrm{dt}$
$\Rightarrow \mathrm{I}_{1}=-\int \mathrm{t}^{\frac{1}{2}} \mathrm{dt}$
Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$
$\Rightarrow \mathrm{I}_{1}=-\frac{\mathrm{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\mathrm{c}$
$\Rightarrow \mathrm{I}_{1}=-\frac{\mathrm{t}^{\frac{3}{2}}}{\frac{3}{2}}+\mathrm{c}$
$\Rightarrow \mathrm{I}_{1}=-\frac{2}{3} \mathrm{t}^{\frac{3}{2}}+\mathrm{c}$
$\Rightarrow \mathrm{I}_{1}=-\frac{2}{3} \mathrm{t}^{\frac{3}{2}}+\mathrm{c}$
$\therefore \mathrm{I}_{1}=-\frac{2}{3}\left(2+3 \mathrm{x}-\mathrm{x}^{2}\right)^{\frac{3}{2}}+\mathrm{c}$
Let $I_{2}=-2 \int \sqrt{2+3 x-x^{2}} d x$
We can write $2+3 x-x^{2}=-\left(x^{2}-3 x-2\right)$
$\Rightarrow 2+3 x-x^{2}=-\left[x^{2}-2(x)\left(\frac{3}{2}\right)+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}-2\right]$
$\Rightarrow 2+3 x-x^{2}=-\left[\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}-2\right]$
$\Rightarrow 2+3 x-x^{2}=-\left[\left(x-\frac{3}{2}\right)^{2}-\frac{17}{4}\right]$
$\Rightarrow 2+3 x-x^{2}=\frac{17}{4}-\left(x-\frac{3}{2}\right)^{2}$
$\Rightarrow 2+3 x-x^{2}=\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}$
Hence, we can write $I_{2}$ as
$I_{2}=-2 \int \sqrt{\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}} d x$
Recall $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c$
$\Rightarrow I_{2}=-2\left[\frac{\left(x-\frac{3}{2}\right)}{2} \sqrt{\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}+\frac{\left(\frac{\sqrt{17}}{2}\right)^{2}}{2} \sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{17}}{2}}\right)\right]+c$
$\Rightarrow \mathrm{I}_{2}=-2\left[\frac{2 \mathrm{x}-3}{4} \sqrt{2+3 \mathrm{x}-\mathrm{x}^{2}}+\frac{17}{8} \sin ^{-1}\left(\frac{2 \mathrm{x}-3}{\sqrt{17}}\right)\right]+\mathrm{c}$
$\therefore \mathrm{I}_{2}=-\frac{1}{2}(2 \mathrm{x}-3) \sqrt{2+3 \mathrm{x}-\mathrm{x}^{2}}-\frac{17}{4} \sin ^{-1}\left(\frac{2 \mathrm{x}-3}{\sqrt{17}}\right)+\mathrm{c}$
Substituting $I_{1}$ and $I_{2}$ in $I$, we get
$I=-\frac{2}{3}\left(2+3 x-x^{2}\right)^{\frac{3}{2}}-\frac{1}{2}(2 x-3) \sqrt{2+3 x-x^{2}}-\frac{17}{4} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{17}}\right)+c$
Thus,
$\int(2 x-5) \sqrt{2+3 x-x^{2}} d x=-\frac{2}{3}\left(2+3 x-x^{2}\right)^{\frac{3}{2}}-\frac{1}{2}(2 x-3) \sqrt{2+3 x-x^{2}}-$
$\frac{17}{4} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{17}}\right)+c$