Question

1. 5 pts. each evaluate the indefinite integrals below using substitution.

• ∫x√(x + 2)dx
• ∫(cos(π/x))/x²dx

Explanation:

First integral: \(\int x\sqrt{x + 2}\,dx\)

Step 1: Substitution

Let \(u=x + 2\), then \(x=u - 2\) and \(dx=du\).
The integral becomes \(\int(u - 2)\sqrt{u}\,du=\int(u-2)u^{\frac{1}{2}}du=\int(u^{\frac{3}{2}}-2u^{\frac{1}{2}})du\)

Step 2: Integrate term - by - term

Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have:
\(\int u^{\frac{3}{2}}du-2\int u^{\frac{1}{2}}du=\frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}-2\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\)
\(=\frac{2}{5}u^{\frac{5}{2}}-\frac{4}{3}u^{\frac{3}{2}}+C\)

Step 3: Back - substitution

Substitute \(u=x + 2\) back in:
\(\frac{2}{5}(x + 2)^{\frac{5}{2}}-\frac{4}{3}(x + 2)^{\frac{3}{2}}+C\)

Second integral: \(\int\frac{\cos(\pi/x)}{x^{2}}dx\)

Step 1: Substitution

Let \(t=\frac{\pi}{x}\), then \(dt=-\frac{\pi}{x^{2}}dx\) and \(\frac{1}{x^{2}}dx=-\frac{1}{\pi}dt\)
The integral becomes \(-\frac{1}{\pi}\int\cos(t)dt\)

Step 2: Integrate

Using the rule \(\int\cos(t)dt=\sin(t)+C\), we get:
\(-\frac{1}{\pi}\sin(t)+C\)

Step 3: Back - substitution

Substitute \(t = \frac{\pi}{x}\) back in:
\(-\frac{1}{\pi}\sin(\frac{\pi}{x})+C\)

Answer:

• For \(\int x\sqrt{x + 2}\,dx\): \(\frac{2}{5}(x + 2)^{\frac{5}{2}}-\frac{4}{3}(x + 2)^{\frac{3}{2}}+C\)
• For \(\int\frac{\cos(\pi/x)}{x^{2}}dx\): \(-\frac{1}{\pi}\sin(\frac{\pi}{x})+C\)