Question

consider the definite integral $int\frac{f(x)}{sqrt{g(x)}}dx$ where $f(x)=x$ and $g(x)=15 - 2x - x^{2}$. (a) rewrite $sqrt{g(x)}$ in the form of $sqrt{b^{2}-(x + a)^{2}}$. (b) use a trigonometric substitution to evaluate the indefinite integral. be sure to draw the right triangle and express your final answer in terms of $x$ as much as possible. you must show all steps. enter your final answer for (a) and (b).

Explanation:

Step1: Complete the square for $g(x)$

We have $g(x)=15 - 2x - x^{2}=-(x^{2}+2x - 15)$. Completing the square for $x^{2}+2x - 15$, we get $x^{2}+2x - 15=(x + 1)^{2}-16$. So $g(x)=16-(x + 1)^{2}$, and $\sqrt{g(x)}=\sqrt{16-(x + 1)^{2}}$, where $a = 1$ and $b = 4$.

Step2: Trigonometric substitution

Let $x+1 = 4\sin\theta$, then $dx=4\cos\theta d\theta$.
The integral $\int\frac{f(x)}{\sqrt{g(x)}}dx=\int\frac{x}{\sqrt{16-(x + 1)^{2}}}dx$. Since $x=4\sin\theta-1$, the integral becomes $\int\frac{4\sin\theta-1}{4\cos\theta}\cdot4\cos\theta d\theta=\int(4\sin\theta - 1)d\theta$.

Step3: Integrate term - by - term

$\int(4\sin\theta - 1)d\theta=-4\cos\theta-\theta + C$.

Step4: Back - substitution

Since $x + 1=4\sin\theta$, then $\sin\theta=\frac{x + 1}{4}$. Using the right - triangle relationship $\sin\theta=\frac{x + 1}{4}$, we have a right - triangle with opposite side $x + 1$, hypotenuse $4$, and adjacent side $\sqrt{16-(x + 1)^{2}}$. So $\cos\theta=\frac{\sqrt{16-(x + 1)^{2}}}{4}$ and $\theta=\arcsin(\frac{x + 1}{4})$.
The integral is $-4\cdot\frac{\sqrt{16-(x + 1)^{2}}}{4}-\arcsin(\frac{x + 1}{4})+C=-\sqrt{16-(x + 1)^{2}}-\arcsin(\frac{x + 1}{4})+C$.

Answer:

(a) $\sqrt{g(x)}=\sqrt{16-(x + 1)^{2}}$
(b) $-\sqrt{16-(x + 1)^{2}}-\arcsin(\frac{x + 1}{4})+C$