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:

Part A:

Step1: Use substitution for integral

Let \( u = \pi v \), then \( du=\pi dv \) or \( dv = \frac{du}{\pi} \). When \( v=\frac{1}{2} \), \( u=\frac{\pi}{2} \); when \( v = 2 \), \( u = 2\pi \).
The integral becomes \( \int_{\frac{\pi}{2}}^{2\pi}\cos(u)\cdot\frac{du}{\pi}=\frac{1}{\pi}\int_{\frac{\pi}{2}}^{2\pi}\cos(u)du \)

Step2: Integrate \( \cos(u) \)

The antiderivative of \( \cos(u) \) is \( \sin(u) \), so \( \frac{1}{\pi}[\sin(u)]_{\frac{\pi}{2}}^{2\pi}=\frac{1}{\pi}(\sin(2\pi)-\sin(\frac{\pi}{2})) \)

Step3: Evaluate the sine values

\( \sin(2\pi) = 0 \), \( \sin(\frac{\pi}{2}) = 1 \), so \( \frac{1}{\pi}(0 - 1)=-\frac{1}{\pi} \)

Part B:

Step1: Substitution for integral

Let \( t = e^{2x}-1 \), then \( dt = 2e^{2x}dx \). When \( x = 0 \), \( t = 0 \); when \( x=\ln 2 \), \( t=e^{2\ln 2}-1 = 4 - 1=3 \)
The integral becomes \( \int_{0}^{3}\sqrt{t}dt \) (since \( e^{2x}dx=\frac{dt}{2} \), and the \( e^{2x} \) in the original integral combines with \( dx \) to give \( \frac{dt}{2} \), and the \( \sqrt{t} \) remains, so the integral is \( \int_{0}^{3}\sqrt{t}dt \))

Step2: Integrate \( \sqrt{t} \)

The antiderivative of \( t^{\frac{1}{2}} \) is \( \frac{2}{3}t^{\frac{3}{2}} \), so \( [\frac{2}{3}t^{\frac{3}{2}}]_{0}^{3}=\frac{2}{3}(3^{\frac{3}{2}}-0) \)

Step3: Simplify \( 3^{\frac{3}{2}} \)

\( 3^{\frac{3}{2}}=3\sqrt{3} \), so \( \frac{2}{3}\times3\sqrt{3}=2\sqrt{3} \)

Part C:

Step1: Substitution for integral

Let \( u = \tan(t) \), then \( du=\sec^{2}(t)dt \). When \( t = 0 \), \( u = 0 \); when \( t=\frac{\pi}{4} \), \( u = 1 \)
The integral becomes \( 2\int_{0}^{1}u^{2}du \) (since \( \sec^{2}(t)dt = du \), and the integral has \( 2\sec^{2}(t)\tan^{2}(t)dt = 2u^{2}du \))

Step2: Integrate \( u^{2} \)

The antiderivative of \( u^{2} \) is \( \frac{1}{3}u^{3} \), so \( 2[\frac{1}{3}u^{3}]_{0}^{1}=2(\frac{1}{3}(1)^{3}-\frac{1}{3}(0)^{3}) \)

Step3: Evaluate the integral

\( 2\times\frac{1}{3}=\frac{2}{3} \)

Part D:

Answer:

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