Evaluate the following integrals –
Question:

Evaluate the following integrals –

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

Solution:

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 $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c$

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