Question

$$ int _ { - 2 } ^ { 0 } \frac { d x } { ( 5 - 4 x - x ^ { 2 } ) ^ { 3 / 2 } } = $$

Explanation:

Step1: Complete the square

$$\begin{align*} 5 - 4x - x^{2}&=-(x^{2}+4x)+5\\ &=-(x^{2}+4x + 4 - 4)+5\\ &=-(x + 2)^{2}+9 \end{align*}$$

So the integral becomes $\int_{-2}^{0}\frac{dx}{(9-(x + 2)^{2})^{\frac{3}{2}}}$.

Step2: Use substitution

Let $x+2 = 3\sin t$, then $dx=3\cos tdt$.
When $x=-2$, $t = 0$; when $x = 0$, $t=\frac{\pi}{6}$.
The integral $\int_{-2}^{0}\frac{dx}{(9-(x + 2)^{2})^{\frac{3}{2}}}=\int_{0}^{\frac{\pi}{6}}\frac{3\cos tdt}{(9 - 9\sin^{2}t)^{\frac{3}{2}}}$.
Since $9-9\sin^{2}t=9\cos^{2}t$, the integral is $\int_{0}^{\frac{\pi}{6}}\frac{3\cos tdt}{(9\cos^{2}t)^{\frac{3}{2}}}=\frac{1}{9}\int_{0}^{\frac{\pi}{6}}\frac{dt}{\cos^{2}t}$.

Step3: Integrate

We know that $\int\frac{dt}{\cos^{2}t}=\tan t+C$.
So $\frac{1}{9}\int_{0}^{\frac{\pi}{6}}\frac{dt}{\cos^{2}t}=\frac{1}{9}[\tan t]_{0}^{\frac{\pi}{6}}$.

Step4: Evaluate the definite - integral

$\frac{1}{9}(\tan\frac{\pi}{6}-\tan0)=\frac{1}{9}(\frac{\sqrt{3}}{3}-0)=\frac{\sqrt{3}}{27}$.

Answer:

$\frac{\sqrt{3}}{27}$