3. find each of the following integrals: (a) ∫(sin⁵(x/3)/√(cos(x/3)))dx (b) ∫(√(x² - 9)/x³)dx (c) ∫(7/(25x²…

Explanation:

(a)

Step1: Rewrite $\sin^{5}(\frac{x}{3})$

We know that $\sin^{5}(\frac{x}{3})=\sin^{4}(\frac{x}{3})\sin(\frac{x}{3})=(1 - \cos^{2}(\frac{x}{3}))^{2}\sin(\frac{x}{3})$. Let $u = \cos(\frac{x}{3})$, then $du=-\frac{1}{3}\sin(\frac{x}{3})dx$ and $\sin(\frac{x}{3})dx=- 3du$. The integral $\int\frac{\sin^{5}(\frac{x}{3})}{\sqrt{\cos(\frac{x}{3})}}dx=\int\frac{(1 - \cos^{2}(\frac{x}{3}))^{2}\sin(\frac{x}{3})}{\sqrt{\cos(\frac{x}{3})}}dx$. Substituting $u = \cos(\frac{x}{3})$ gives $-3\int\frac{(1 - u^{2})^{2}}{\sqrt{u}}du=-3\int\frac{1 - 2u^{2}+u^{4}}{\sqrt{u}}du=-3\int(u^{-\frac{1}{2}}-2u^{\frac{3}{2}}+u^{\frac{7}{2}})du$.

Step2: Integrate term - by - term

$-3\left(2u^{\frac{1}{2}}-\frac{4}{5}u^{\frac{5}{2}}+\frac{2}{9}u^{\frac{9}{2}}\right)+C=-6\sqrt{u}+\frac{12}{5}u^{\frac{5}{2}}-\frac{2}{3}u^{\frac{9}{2}}+C$. Substituting back $u = \cos(\frac{x}{3})$, we get $-6\sqrt{\cos(\frac{x}{3})}+\frac{12}{5}\cos^{\frac{5}{2}}(\frac{x}{3})-\frac{2}{3}\cos^{\frac{9}{2}}(\frac{x}{3})+C$.

(b)

Step1: Use the trigonometric substitution

Let $x = 3\sec\theta$, then $dx=3\sec\theta\tan\theta d\theta$ and $\sqrt{x^{2}-9}=3\tan\theta$. The integral $\int\frac{\sqrt{x^{2}-9}}{x^{3}}dx=\int\frac{3\tan\theta}{27\sec^{3}\theta}\cdot3\sec\theta\tan\theta d\theta=\frac{1}{3}\int\frac{\tan^{2}\theta}{\sec^{2}\theta}d\theta=\frac{1}{3}\int\sin^{2}\theta d\theta$.

Step2: Use the double - angle formula

Since $\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}$, then $\frac{1}{3}\int\sin^{2}\theta d\theta=\frac{1}{6}\int(1 - \cos(2\theta))d\theta=\frac{1}{6}(\theta-\frac{1}{2}\sin(2\theta))+C$. Since $x = 3\sec\theta$, $\theta=\text{arcsec}(\frac{x}{3})$ and $\sin\theta=\frac{\sqrt{x^{2}-9}}{x}$, $\cos\theta=\frac{3}{x}$. $\sin(2\theta)=2\sin\theta\cos\theta=\frac{6\sqrt{x^{2}-9}}{x^{2}}$. So the integral is $\frac{1}{6}\text{arcsec}(\frac{x}{3})-\frac{\sqrt{x^{2}-9}}{2x^{2}}+C$.

(c)

Step1: Complete the square in the denominator

$25x^{2}-20x + 13=25\left(x^{2}-\frac{4}{5}x\right)+13=25\left(x^{2}-\frac{4}{5}x+\frac{4}{25}-\frac{4}{25}\right)+13=25\left((x - \frac{2}{5})^{2}+\frac{9}{25}\right)=25\left(x-\frac{2}{5}\right)^{2}+9$. Let $u=x-\frac{2}{5}$, then $du = dx$ and the integral $\int\frac{7}{25x^{2}-20x + 13}dx=\int\frac{7}{25u^{2}+9}du$.

Step2: Use the arctangent formula

$\int\frac{7}{25u^{2}+9}du=\frac{7}{25}\int\frac{1}{u^{2}+\frac{9}{25}}du$. Since $\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\arctan(\frac{x}{a})+C$ with $a=\frac{3}{5}$, we have $\frac{7}{25}\cdot\frac{5}{3}\arctan(\frac{5u}{3})+C=\frac{7}{15}\arctan(\frac{5x - 2}{3})+C$.

(d)

Step1: Use partial fraction decomposition

$\frac{5x^{2}}{(x + 2)(x^{2}+1)}=\frac{A}{x + 2}+\frac{Bx+C}{x^{2}+1}$. Then $5x^{2}=A(x^{2}+1)+(Bx + C)(x + 2)$. Let $x=-2$, we get $20 = 5A$, so $A = 4$. Expanding and comparing coefficients: $5x^{2}=A x^{2}+A + Bx^{2}+2Bx + Cx+2C=(A + B)x^{2}+(2B + C)x+(A + 2C)$. Since $A = 4$, then $A + B=5$ gives $B = 1$, and $2B + C=0$ gives $C=-2$. So $\frac{5x^{2}}{(x + 2)(x^{2}+1)}=\frac{4}{x + 2}+\frac{x - 2}{x^{2}+1}=\frac{4}{x + 2}+\frac{x}{x^{2}+1}-\frac{2}{x^{2}+1}$.

Step2: Integrate term - by - term

$\int\frac{5x^{2}}{(x + 2)(x^{2}+1)}dx=4\int\frac{1}{x + 2}dx+\frac{1}{2}\int\frac{2x}{x^{2}+1}dx-2\int\frac{1}{x^{2}+1}dx=4\ln|x + 2|+\frac{1}{2}\ln(x^{2}+1)-2\arctan(x)+C$.

(e)

Step1: Use integration by parts

The formula for integration by parts is $\int u dv=uv-\int v du$. Let $u = x$ and $dv=e^{x}dx$, then $du = dx$ and $v = e^{x}$. So $\int xe^{x}dx=xe^{x}-\int e^{x}dx$.

Step2: Evaluate the definite integral

$\int_{-\infty}^{0}xe^{x}dx=\lim_{a\rightarrow-\infty}\int_{a}^{0}xe^{x}dx=\lim_{a\rightarrow-\infty}\left[xe^{x}-e^{x}\right]{a}^{0}=\lim{a\rightarrow-\infty}\left((0 - 1)-(ae^{a}-e^{a})\right)$. Using L'Hopital's rule on $\lim_{a\rightarrow-\infty}ae^{a}=\lim_{a\rightarrow-\infty}\frac{a}{e^{-a}}=\lim_{a\rightarrow-\infty}\frac{1}{-e^{-a}} = 0$. So $\int_{-\infty}^{0}xe^{x}dx=-1$.

Answer:

(a) $-6\sqrt{\cos(\frac{x}{3})}+\frac{12}{5}\cos^{\frac{5}{2}}(\frac{x}{3})-\frac{2}{3}\cos^{\frac{9}{2}}(\frac{x}{3})+C$ (b) $\frac{1}{6}\text{arcsec}(\frac{x}{3})-\frac{\sqrt{x^{2}-9}}{2x^{2}}+C$ (c) $\frac{7}{15}\arctan(\frac{5x - 2}{3})+C$ (d) $4\ln|x + 2|+\frac{1}{2}\ln(x^{2}+1)-2\arctan(x)+C$ (e) $-1$