# Evaluate the following integrals -

Question:

Evaluate the following integrals -

$\int(2 x+5) \sqrt{10-4 x-3 x^{2}} d x$

Solution:

Let $I=\int(2 x+5) \sqrt{10-4 x-3 x^{2}} d x$

Let us assume, $2 \mathrm{x}+5=\lambda \frac{\mathrm{d}}{\mathrm{dx}}\left(10-4 \mathrm{x}-3 \mathrm{x}^{2}\right)+\mu$

$\Rightarrow 2 x+5=\lambda\left[\frac{d}{d x}(10)-\frac{d}{d x}(4 x)-\frac{d}{d x}\left(3 x^{2}\right)\right]+\mu$

$\Rightarrow 2 x+5=\lambda\left[\frac{d}{d x}(10)-4 \frac{d}{d x}(x)-3 \frac{d}{d x}\left(x^{2}\right)\right]+\mu$

We know $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$ and derivative of a constant is 0 .

$\Rightarrow 2 x+5=\lambda\left(0-4-3 \times 2 x^{2-1}\right)+\mu$

$\Rightarrow 2 x+5=\lambda(-4-6 x)+\mu$

$\Rightarrow 2 x+5=-6 \lambda x+\mu-4 \lambda$

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

$-6 \lambda=2 \Rightarrow \lambda=-\frac{2}{6}=-\frac{1}{3}$

Comparing the constant on both sides, we get

$\mu-4 \lambda=5$

$\Rightarrow \mu-4\left(-\frac{1}{3}\right)=5$

$\Rightarrow \mu+\frac{4}{3}=5$

$\therefore \mu=\frac{11}{3}$]

Hence, we have $2 x+5=-\frac{1}{3}(-4-6 x)+\frac{11}{3}$

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

$I=\int\left[-\frac{1}{3}(-4-6 x)+\frac{11}{3}\right] \sqrt{10-4 x-3 x^{2}} d x$

$\Rightarrow \mathrm{I}=\int\left[-\frac{1}{3}(-4-6 \mathrm{x}) \sqrt{10-4 \mathrm{x}-3 \mathrm{x}^{2}}+\frac{11}{3} \sqrt{10-4 \mathrm{x}-3 \mathrm{x}^{2}}\right] \mathrm{dx}$

$\Rightarrow \mathrm{I}=-\int \frac{1}{3}(-4-6 \mathrm{x}) \sqrt{10-4 \mathrm{x}-3 \mathrm{x}^{2}} \mathrm{dx}+\int \frac{11}{3} \sqrt{10-4 \mathrm{x}-3 \mathrm{x}^{2}} \mathrm{dx}$

$\Rightarrow \mathrm{I}=-\frac{1}{3} \int(-4-6 \mathrm{x}) \sqrt{10-4 \mathrm{x}-3 \mathrm{x}^{2}} \mathrm{dx}+\frac{11}{3} \int \sqrt{10-4 \mathrm{x}-3 \mathrm{x}^{2}} \mathrm{dx}$

Let $I_{1}=-\frac{1}{3} \int(-4-6 x) \sqrt{10-4 x-3 x^{2}} d x$

Now, put $10-4 x-3 x^{2}=t$

$\Rightarrow(-4-6 x) d x=d t$ (Differentiating both sides)

Substituting this value in $\mathrm{I}_{1}$, we can write

$\mathrm{I}_{1}=-\frac{1}{3} \int \sqrt{\mathrm{t} \mathrm{d} \mathrm{t}}$

$\Rightarrow \mathrm{I}_{1}=-\frac{1}{3} \int \mathrm{t} \frac{1}{2} \mathrm{dt}$

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

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

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

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

$\Rightarrow \mathrm{I}_{1}=-\frac{2}{9} \mathrm{t} \frac{3}{2}+\mathrm{c}$

$\therefore \mathrm{I}_{1}=-\frac{2}{9}\left(10-4 \mathrm{x}-3 \mathrm{x}^{2}\right)^{\frac{3}{2}}+\mathrm{c}$

Let $I_{2}=\frac{11}{3} \int \sqrt{10-4 x-3 x^{2}} d x$

We can write $10-4 x-3 x^{2}=-\left(3 x^{2}+4 x-10\right)$

$\Rightarrow 10-4 x-3 x^{2}=-3\left[x^{2}+\frac{4}{3} x-\frac{10}{3}\right]$

$\Rightarrow 10-4 \mathrm{x}-3 \mathrm{x}^{2}=-3\left[\mathrm{x}^{2}+2(\mathrm{x})\left(\frac{2}{3}\right)+\left(\frac{2}{3}\right)^{2}-\left(\frac{2}{3}\right)^{2}-\frac{10}{3}\right]$

$\Rightarrow 10-4 \mathrm{x}-3 \mathrm{x}^{2}=-3\left[\left(\mathrm{x}+\frac{2}{3}\right)^{2}-\frac{4}{9}-\frac{10}{3}\right]$

$\Rightarrow 10-4 x-3 x^{2}=-3\left[\left(x+\frac{2}{3}\right)^{2}-\frac{34}{9}\right]$

$\Rightarrow 10-4 x-3 x^{2}=3\left[\frac{34}{9}-\left(x+\frac{2}{3}\right)^{2}\right]$

$\Rightarrow 10-4 x-3 x^{2}=3\left[\left(\frac{\sqrt{34}}{3}\right)^{2}-\left(x+\frac{2}{3}\right)^{2}\right]$

Hence, we can write $\mathrm{I}_{2}$ as

$\mathrm{I}_{2}=\frac{11}{3} \int \sqrt{3\left[\left(\frac{\sqrt{34}}{3}\right)^{2}-\left(\mathrm{x}+\frac{2}{3}\right)^{2}\right]} \mathrm{dx}$

$\Rightarrow \mathrm{I}_{2}=\frac{11 \sqrt{3}}{3} \int \sqrt{\left(\frac{\sqrt{34}}{3}\right)^{2}-\left(\mathrm{x}+\frac{2}{3}\right)^{2}} \mathrm{dx}$

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

$\Rightarrow \mathrm{I}_{2}=\frac{11 \sqrt{3}}{3}\left[\frac{\left(\mathrm{x}+\frac{2}{3}\right)}{2} \sqrt{\left(\frac{\sqrt{34}}{3}\right)^{2}-\left(\mathrm{x}+\frac{2}{3}\right)^{2}}+\frac{\left(\frac{\sqrt{34}}{3}\right)^{2}}{2} \sin ^{-1}\left(\frac{\mathrm{x}+\frac{2}{3}}{\frac{\sqrt{34}}{3}}\right)\right]+\mathrm{c}$

$\Rightarrow \mathrm{I}_{2}=\frac{11 \sqrt{3}}{3}\left[\frac{(3 \mathrm{x}+2)}{6} \sqrt{\frac{10}{3}-\frac{4}{3} \mathrm{x}-\mathrm{x}^{2}+\frac{34}{18}} \sin ^{-1}\left(\frac{3 \mathrm{x}+2}{\sqrt{34}}\right)\right]+\mathrm{c}$

$\Rightarrow I_{2}=-\frac{11 \sqrt{3}}{18}(3 x+2) \sqrt{\frac{10}{3}-\frac{4}{3} x-x^{2}}-\frac{374 \sqrt{3}}{54} \sin ^{-1}\left(\frac{3 x+2}{\sqrt{34}}\right)+c$

$\therefore I_{2}=-\frac{11}{18}(3 x+2) \sqrt{10-4 x-3 x^{2}}-\frac{187 \sqrt{3}}{27} \sin ^{-1}\left(\frac{3 x+2}{\sqrt{34}}\right)+c$

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

$I=-\frac{2}{9}\left(10-4 x-3 x^{2}\right)^{\frac{3}{2}}-\frac{11}{18}(3 x+2) \sqrt{10-4 x-3 x^{2}}$

$-\frac{187 \sqrt{3}}{27} \sin ^{-1}\left(\frac{3 x+2}{\sqrt{34}}\right)+c$

Thus,$\int(2 x+5) \sqrt{10-4 x-3 x^{2}} d x=-\frac{2}{9}\left(10-4 x-3 x^{2}\right)^{\frac{3}{2}}-\frac{11}{18}(3 x+$

2) $\sqrt{10-4 x-3 x^{2}}-\frac{187 \sqrt{3}}{27} \sin ^{-1}\left(\frac{3 x+2}{\sqrt{34}}\right)+c$