Evaluate the following integrals -

Solution:

Let $I=\int(x+3) \sqrt{3-4 x-x^{2}} d x$

Let us assume $x+3=\lambda \frac{d}{d x}\left(3-4 x-x^{2}\right)+\mu$

$\Rightarrow \mathrm{x}+3=\lambda\left[\frac{\mathrm{d}}{\mathrm{dx}}(3)-\frac{\mathrm{d}}{\mathrm{dx}}(4 \mathrm{x})-\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)\right]+\mu$

$\Rightarrow \mathrm{x}+3=\lambda\left[\frac{\mathrm{d}}{\mathrm{dx}}(3)-4 \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)\right]+\mu$

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 .

$\Rightarrow x+3=\lambda\left(0-4-2 x^{2-1}\right)+\mu$

$\Rightarrow x+3=\lambda(-4-2 x)+\mu$

$\Rightarrow x+3=-2 \lambda x+\mu-4 \lambda$

Comparing the coefficient of $x$ on both sides, we get

$-2 \lambda=1 \Rightarrow \lambda=-\frac{1}{2}$

Comparing the constant on both sides, we get

$\mu-4 \lambda=3$

$\Rightarrow \mu-4\left(-\frac{1}{2}\right)=3$

$\Rightarrow \mu+2=3$

$\therefore \mu=1$

Hence, we have $x+3=-\frac{1}{2}(-4-2 x)+1$

Substituting this value in I, we can write the integral as

$I=\int\left[-\frac{1}{2}(-4-2 x)+1\right] \sqrt{3-4 x-x^{2}} d x$

$\Rightarrow I=\int\left[-\frac{1}{2}(-4-2 x) \sqrt{3-4 x-x^{2}}+\sqrt{3-4 x-x^{2}}\right] d x$

$\Rightarrow I=-\int \frac{1}{2}(-4-2 x) \sqrt{3-4 x-x^{2}} d x+\int \sqrt{3-4 x-x^{2}} d x$

$\Rightarrow I=-\frac{1}{2} \int(-4-2 x) \sqrt{3-4 x-x^{2}} d x+\int \sqrt{3-4 x-x^{2}} d x$

Let $\mathrm{I}_{1}=-\frac{1}{2} \int(-4-2 \mathrm{x}) \sqrt{3-4 \mathrm{x}-\mathrm{x}^{2}} \mathrm{dx}$

Now, put $3-4 x-x^{2}=t$

$\Rightarrow(-4-2 x) d x=d t$ (Differentiating both sides)

Substituting this value in $I_{1}$, we can write

$\mathrm{I}_{1}=-\frac{1}{2} \int \sqrt{\mathrm{t} \mathrm{d} \mathrm{t}}$

$\Rightarrow \mathrm{I}_{1}=-\frac{1}{2} \int \mathrm{t}^{\frac{1}{2}} \mathrm{dt}$

Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow I_{1}=-\frac{1}{2}\left(\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)+c$

$\Rightarrow \mathrm{I}_{1}=-\frac{1}{2}\left(\frac{\mathrm{t}^{\frac{3}{2}}}{\frac{3}{2}}\right)+\mathrm{c}$

$\Rightarrow \mathrm{I}_{1}=-\frac{1}{2} \times \frac{2}{3} \mathrm{t} \frac{3}{2}+\mathrm{c}$

$\Rightarrow \mathrm{I}_{1}=-\frac{1}{3} \mathrm{t}^{\frac{3}{2}}+\mathrm{c}$

$\therefore \mathrm{I}_{1}=-\frac{1}{3}\left(3-4 \mathrm{x}-\mathrm{x}^{2}\right)^{\frac{3}{2}}+\mathrm{c}$

Let $I_{2}=\int \sqrt{3-4 x-x^{2}} d x$

We can write $3-4 x-x^{2}=-\left(x^{2}+4 x-3\right)$

$\Rightarrow 3-4 x-x^{2}=-\left[x^{2}+2(x)(2)+2^{2}-2^{2}-3\right]$

$\Rightarrow 3-4 x-x^{2}=-\left[(x+2)^{2}-4-3\right]$

$\Rightarrow 3-4 x-x^{2}=-\left[(x+2)^{2}-7\right]$

$\Rightarrow 3-4 x-x^{2}=7-(x+2)^{2}$

$\Rightarrow 3-4 x-x^{2}=(\sqrt{7})^{2}-(x+2)^{2}$

Hence, we can write $I_{2}$ as

$I_{2}=\int \sqrt{(\sqrt{7})^{2}-(x+2)^{2}} d x$

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

$\Rightarrow I_{2}=\frac{(x+2)}{2} \sqrt{(\sqrt{7})^{2}-(x+2)^{2}}+\frac{(\sqrt{7})^{2}}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+c$

$\therefore \mathrm{I}_{2}=\frac{1}{2}(\mathrm{x}+2) \sqrt{3-4 \mathrm{x}-\mathrm{x}^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{\mathrm{x}+2}{\sqrt{7}}\right)+\mathrm{c}$

Substituting $I_{1}$ and $I_{2}$ in $I$, we get

$I=-\frac{1}{3}\left(3-4 x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}(x+2) \sqrt{3-4 x-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+c$

Thus, $\int(x+3) \sqrt{3-4 x-x^{2}} d x=-\frac{1}{3}\left(3-4 x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}(x+2) \sqrt{3-4 x-x^{2}}+$

$\frac{7}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+c$