Question

evaluate the integral by completing the square and using trigonometric substitution. (int\frac{dx}{sqrt{3 + x - x^{2}}}) (express numbers in exact form. use symbolic notation and fractions where needed. use (c) for into (c) as much as possible.) (int\frac{dx}{sqrt{3 + x - x^{2}}}=)

Explanation:

Step1: Complete the square for the quadratic

First, rewrite $3 + x - x^{2}$ as $-(x^{2}-x - 3)$. Completing the square for $x^{2}-x - 3$:
\[

$$\begin{align*} x^{2}-x - 3&=(x-\frac{1}{2})^{2}-\frac{1}{4}-3\\ &=(x - \frac{1}{2})^{2}-\frac{13}{4} \end{align*}$$

\]
So, $3 + x - x^{2}=\frac{13}{4}-(x-\frac{1}{2})^{2}$.

Step2: Make a trigonometric substitution

Let $u=x-\frac{1}{2}$, then $du = dx$ and the integral becomes $\int\frac{du}{\sqrt{\frac{13}{4}-u^{2}}}$.
Let $u=\frac{\sqrt{13}}{2}\sin\theta$, then $du=\frac{\sqrt{13}}{2}\cos\theta d\theta$.
\[

$$\begin{align*} \sqrt{\frac{13}{4}-u^{2}}&=\sqrt{\frac{13}{4}-\frac{13}{4}\sin^{2}\theta}\\ &=\frac{\sqrt{13}}{2}\cos\theta \end{align*}$$

\]
The integral $\int\frac{du}{\sqrt{\frac{13}{4}-u^{2}}}=\int\frac{\frac{\sqrt{13}}{2}\cos\theta d\theta}{\frac{\sqrt{13}}{2}\cos\theta}=\int d\theta=\theta + C$.

Step3: Back - substitute

Since $u=\frac{\sqrt{13}}{2}\sin\theta$, then $\sin\theta=\frac{2u}{\sqrt{13}}$ and $\theta=\arcsin(\frac{2u}{\sqrt{13}})$.
Substituting back $u = x-\frac{1}{2}$, we get $\arcsin(\frac{2(x - \frac{1}{2})}{\sqrt{13}})+C=\arcsin(\frac{2x - 1}{\sqrt{13}})+C$.

Answer:

$\arcsin(\frac{2x - 1}{\sqrt{13}})+C$