Question

1. find the following definite integrals.

a. \\( \int_{1/2}^{2} \cos(\pi v) \\, dv \\) c. \\( \int_{0}^{\pi/4} 2\sec^2(t)\tan^2 t \\, dt \\)
b. \\( \int_{0}^{\ln 2} e^{2x} \sqrt{e^{2x} - 1} \\, dx \\) d. \\( \int_{0}^{\pi/4} \tan x \\, dx \\)

Explanation:

Step1: Solve integral A

Let \(u = \pi v\), then \(du=\pi dv\), \(dv=\frac{du}{\pi}\).
When \(v = \frac{1}{2}\), \(u=\frac{\pi}{2}\); when \(v = 2\), \(u = 2\pi\).
\(\int_{\frac{1}{2}}^{2}\cos(\pi v)dv=\frac{1}{\pi}\int_{\frac{\pi}{2}}^{2\pi}\cos(u)du\)
\(=\frac{1}{\pi}[\sin(u)]_{\frac{\pi}{2}}^{2\pi}\)
\(=\frac{1}{\pi}(\sin(2\pi)-\sin(\frac{\pi}{2}))\)
\(=\frac{1}{\pi}(0 - 1)=-\frac{1}{\pi}\)

Step2: Solve integral B

Let \(t=e^{2x}-1\), then \(dt = 2e^{2x}dx\), \(e^{2x}dx=\frac{dt}{2}\).
When \(x = 0\), \(t=0\); when \(x=\ln2\), \(t=e^{2\ln2}-1=4 - 1 = 3\).
\(\int_{0}^{\ln2}e^{2x}\sqrt{e^{2x}-1}dx=\frac{1}{2}\int_{0}^{3}\sqrt{t}dt\)
\(=\frac{1}{2}\times\frac{2}{3}[t^{\frac{3}{2}}]_{0}^{3}\)
\(=\frac{1}{3}(3^{\frac{3}{2}}-0)=\sqrt{3}\)

Step3: Solve integral C

Let \(u=\tan(t)\), then \(du=\sec^{2}(t)dt\).
When \(t = 0\), \(u = 0\); when \(t=\frac{\pi}{4}\), \(u = 1\).
\(\int_{0}^{\frac{\pi}{4}}2\sec^{2}(t)\tan^{2}(t)dt=2\int_{0}^{1}u^{2}du\)
\(=2\times\frac{1}{3}[u^{3}]_{0}^{1}\)
\(=\frac{2}{3}(1 - 0)=\frac{2}{3}\)

Step4: Solve integral D

\(\int_{0}^{\frac{\pi}{4}}\tan(x)dx=\int_{0}^{\frac{\pi}{4}}\frac{\sin(x)}{\cos(x)}dx\)
Let \(u=\cos(x)\), then \(du=-\sin(x)dx\), \(\sin(x)dx=-du\).
When \(x = 0\), \(u = 1\); when \(x=\frac{\pi}{4}\), \(u=\frac{\sqrt{2}}{2}\).
\(\int_{0}^{\frac{\pi}{4}}\frac{\sin(x)}{\cos(x)}dx=-\int_{1}^{\frac{\sqrt{2}}{2}}\frac{1}{u}du\)
\(=-[\ln|u|]_{1}^{\frac{\sqrt{2}}{2}}\)
\(=-(\ln(\frac{\sqrt{2}}{2})-\ln(1))\)
\(=-(\ln(\sqrt{2})-\ln(2)-0)\)
\(=-(\frac{1}{2}\ln(2)-\ln(2))=\frac{1}{2}\ln(2)\)

Answer:

A. \(-\frac{1}{\pi}\)
B. \(\sqrt{3}\)
C. \(\frac{2}{3}\)
D. \(\frac{1}{2}\ln(2)\)