hw091221 - sol091221 - 《Advanced maths (a liberal arts course)》

09-December-2021 Homework:

Openstax Calculus textbook (volume 1):

Page 592, Exercises 5.5, Problems 306, 307, 309, 311, 312.

Page 605, Exercises 5.6, Problems 331, 334, 344, 346, 349, 355, 358, 372.

Solutions

Exercises 5.5

Prob. 306. Let $sol091221 - 图1$ #card=math&code=u%20%3D%20%5Ccos%20%282%5Ctheta%29). Then $sol091221 - 图2$ %5C%2C%20d%5Ctheta#card=math&code=d%20u%20%3D%20-2%5Csin%282%5Ctheta%29%5C%2C%20d%5Ctheta). We have

$sol091221 - 图3$ %20%5Csin%20(2%5Ctheta)%5C%2C%20d%5Ctheta%20%3D%20-%5Cfrac%7B1%7D%7B6%7D%20%5Cint_0%5E%7B%5Cpi%7D%203%5Ccos%5E2%20(2%5Ctheta)%20d%20(%5Ccos%202%5Ctheta)%20%3D%20-%5Cfrac%7B1%7D%7B6%7D%20%5Cint_1%5E1%20d%20(%5Ccos%5E3%20(2%5Ctheta)%20)%20%3D%200#card=math&code=%5Cint_0%5E%7B%5Cpi%7D%20%5Ccos%5E2%282%5Ctheta%29%20%5Csin%20%282%5Ctheta%29%5C%2C%20d%5Ctheta%20%3D%20-%5Cfrac%7B1%7D%7B6%7D%20%5Cint_0%5E%7B%5Cpi%7D%203%5Ccos%5E2%20%282%5Ctheta%29%20d%20%28%5Ccos%202%5Ctheta%29%20%3D%20-%5Cfrac%7B1%7D%7B6%7D%20%5Cint_1%5E1%20d%20%28%5Ccos%5E3%20%282%5Ctheta%29%20%29%20%3D%200).

Prob. 307. Let $sol091221 - 图4$ . Then $sol091221 - 图5$ .

$sol091221 - 图6$ %20%5Csin(t%5E2)%20%5C%2C%20dt%20%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5Cint_0%5E%7B%5Csqrt%7B%5Cpi%7D%7D2%5Csin%20(t%5E2)%20%5Ccos%20(t%5E2)%20d(t%5E2)%20%3D%20%5Cfrac%7B1%7D%7B4%7D%5Cint_0%5E%7B%5Cpi%7D%202%5Csin(u)d%20(%5Csin(u))%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5Cint_0%5E%7B%5Cpi%7D%20d%20(%5Csin%5E2(u))%20%3D0#card=math&code=%5Cint_0%5E%7B%5Csqrt%7B%5Cpi%7D%7D%20t%20%5Ccos%28t%5E2%29%20%5Csin%28t%5E2%29%20%5C%2C%20dt%20%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5Cint_0%5E%7B%5Csqrt%7B%5Cpi%7D%7D2%5Csin%20%28t%5E2%29%20%5Ccos%20%28t%5E2%29%20d%28t%5E2%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D%5Cint_0%5E%7B%5Cpi%7D%202%5Csin%28u%29d%20%28%5Csin%28u%29%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5Cint_0%5E%7B%5Cpi%7D%20d%20%28%5Csin%5E2%28u%29%29%20%3D0) , since $sol091221 - 图7$ %3D%5Csin(0)%3D0#card=math&code=%5Csin%28%5Cpi%29%3D%5Csin%280%29%3D0).

Prob. 309. Let $sol091221 - 图8$ , then $sol091221 - 图9$ . As $sol091221 - 图10$ goes from $sol091221 - 图11$ to $sol091221 - 图12$ , $sol091221 - 图13$ will go from $sol091221 - 图14$ to $sol091221 - 图15$ .

$sol091221 - 图16$ %5E2%7D%20dt%20%3D%20%5Cint%7B-1%2F2%7D%5E%7B1%2F2%7D%20%5Cfrac%7B-2u%7D%7B1%2Bu%5E2%7D%20du%20%3D0#card=math&code=%5Cint_0%5E1%20%5Cfrac%7B1-2t%7D%7B1%2B%28t-%5Cfrac%7B1%7D%7B2%7D%29%5E2%7D%20dt%20%3D%20%5Cint%7B-1%2F2%7D%5E%7B1%2F2%7D%20%5Cfrac%7B-2u%7D%7B1%2Bu%5E2%7D%20du%20%3D0), since the integrand $sol091221 - 图17$ is an odd function with respect to $sol091221 - 图18$ , and the interval $sol091221 - 图19$ is symmetric about the origin.

Prob. 311. Let $sol091221 - 图20$ , then $sol091221 - 图21$ . As $sol091221 - 图22$ goes from $sol091221 - 图23$ to $sol091221 - 图24$ , $sol091221 - 图25$ will go from $sol091221 - 图26$ to $sol091221 - 图27$ .

$sol091221 - 图28$ %20%5Ccos%20(%5Cpi%20t)%5C%2C%20dt%20%3D%20%5Cint1%5E%7B-1%7D%20u%5C%2C%20%5Ccos%20(%5Cpi%20-%20%5Cpi%20u))%20(-du)%20%3D%20-%5Cint%7B-1%7D%5E1%20u%20%5Ccos%20(%5Cpi%20u)%20%5C%2C%20du%20%3D0#card=math&code=%5Cint0%5E2%20%281-t%29%20%5Ccos%20%28%5Cpi%20t%29%5C%2C%20dt%20%3D%20%5Cint_1%5E%7B-1%7D%20u%5C%2C%20%5Ccos%20%28%5Cpi%20-%20%5Cpi%20u%29%29%20%28-du%29%20%3D%20-%5Cint%7B-1%7D%5E1%20u%20%5Ccos%20%28%5Cpi%20u%29%20%5C%2C%20du%20%3D0), because $sol091221 - 图29$ #card=math&code=u%20%5Ccos%20%28%5Cpi%20u%29) is odd, and $sol091221 - 图30$ is symmetric about the origin.

Prob. 312. Let $sol091221 - 图31$ . We have

$sol091221 - 图32$ , because the integrand $\cos^2 u \sin u $ is odd, and $sol091221 - 图33$ is symmetric about the origin.

Exercises 5.6

Prob. 331. Let $sol091221 - 图34$ . Then $sol091221 - 图35$ . Thus, $sol091221 - 图36$ %7D%20%3D%20%5Cint%20%5Cfrac%7Bd%20u%7D%7Bu%20%5Cln%20u%7D%20%3D%20%5Cint%20%5Cfrac%7Bd%20(%5Cln%20u)%7D%7B%5Cln%20u%7D%20%3D%20%5Cln%20(%5Cln%20u)%2BC%20%3D%20%5Cln%20(%5Cln%20(%5Cln%20x))%2BC#card=math&code=%5Cint%20%5Cfrac%7Bd%20x%7D%7Bx%20%5Cln%20x%20%5Cln%28%5Cln%20x%29%7D%20%3D%20%5Cint%20%5Cfrac%7Bd%20u%7D%7Bu%20%5Cln%20u%7D%20%3D%20%5Cint%20%5Cfrac%7Bd%20%28%5Cln%20u%29%7D%7B%5Cln%20u%7D%20%3D%20%5Cln%20%28%5Cln%20u%29%2BC%20%3D%20%5Cln%20%28%5Cln%20%28%5Cln%20x%29%29%2BC).

Prob. 334. Since $sol091221 - 图37$ )%20%3D%20%5Cfrac%7B%5Ccos%20x%7D%7B%5Csin%20x%7D%20dx%20%3D%20%5Cfrac%7Bd%20x%7D%7B%5Ctan%20x%7D#card=math&code=d%28%5Cln%20%28%5Csin%20x%29%29%20%3D%20%5Cfrac%7B%5Ccos%20x%7D%7B%5Csin%20x%7D%20dx%20%3D%20%5Cfrac%7Bd%20x%7D%7B%5Ctan%20x%7D), so

$sol091221 - 图38$ %7D%7B%5Ctan%20x%7Dd%20x%20%3D%20%5Cint%20%5Cln(%5Csin(x))%20d%20(%5Cln(%5Csin%20x))%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cln%5E2%20(%5Csin%20x)%20%2BC#card=math&code=%5Cint%20%5Cfrac%7B%5Cln%20%28%5Csin%20x%29%7D%7B%5Ctan%20x%7Dd%20x%20%3D%20%5Cint%20%5Cln%28%5Csin%28x%29%29%20d%20%28%5Cln%28%5Csin%20x%29%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cln%5E2%20%28%5Csin%20x%29%20%2BC).

Prob. 344. Let $sol091221 - 图39$ , then $sol091221 - 图40$ %2BC#card=math&code=%5Cint%20%5Cfrac%7B%5Cln%20x%7D%7Bx%5E2%7D%20dx%20%3D%20%5Cint%20%5Cln%20u%20%5C%2C%20du%20%3D%20u%5Cln%20u%20-%20u%20%2BC%20%3D%20-%5Cfrac%7B1%7D%7Bx%7D%28%5Cln%20x%20%2B%201%29%2BC).

Prob. 346. The integral is $sol091221 - 图41$ . (Note: because $sol091221 - 图42$ for all $sol091221 - 图43$ , and when $sol091221 - 图44$ , $sol091221 - 图45$ . Thus, the area under the graph of $sol091221 - 图46$ from $sol091221 - 图47$ to $sol091221 - 图48$ is always positive. Thus, we should use the absolute value of $sol091221 - 图49$ , but not just $sol091221 - 图50$ itself.)

Prob. 349. Because $sol091221 - 图51$ %2B%5Ccos(3x)%5D%20%5Cbig)’%20%3D%20%5Cfrac%7B3%5Ccos(3x)-3%5Csin(3x)%7D%7B%5Csin(3x)%2B%5Ccos(3x)%7D#card=math&code=%5Cbig%28%5Cln%20%5B%5Csin%283x%29%2B%5Ccos%283x%29%5D%20%5Cbig%29%27%20%3D%20%5Cfrac%7B3%5Ccos%283x%29-3%5Csin%283x%29%7D%7B%5Csin%283x%29%2B%5Ccos%283x%29%7D), the original integral is equal to

$sol091221 - 图52$ %2B%5Ccos(3x)%5D%20)%20%3D%20-%5Cfrac%7B1%7D%7B3%7D%20%5Cln%20%5B%5Csin(3x)%2B%5Ccos(3x)%5D%20%2BC#card=math&code=-%5Cfrac%7B1%7D%7B3%7D%20%5Cint%20d%28%5Cln%20%5B%5Csin%283x%29%2B%5Ccos%283x%29%5D%20%29%20%3D%20-%5Cfrac%7B1%7D%7B3%7D%20%5Cln%20%5B%5Csin%283x%29%2B%5Ccos%283x%29%5D%20%2BC).

Prob. 355. Because $sol091221 - 图53$ ‘%3D%203%2B6x%2B3x%5E2%20%3D%203(1%2B2x%2Bx%5E2)#card=math&code=%283x%2B3x%5E2%2Bx%5E3%29%27%3D%203%2B6x%2B3x%5E2%20%3D%203%281%2B2x%2Bx%5E2%29), the original integral is equal to $sol091221 - 图54$ %20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cln%20(%5Cfrac%7B26%7D%7B7%7D)#card=math&code=%5Cfrac%7B1%7D%7B3%7D%20%5Cint_1%5E2%20d%28%5Cln%20%7C3x%2B3x%5E2%2Bx%5E3%7C%29%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cln%20%28%5Cfrac%7B26%7D%7B7%7D%29).

Prob. 358. Using Formula 14 in Appendix A (basic integrals), we have

$sol091221 - 图55$ #card=math&code=%5Cint%7B%5Cpi%2F6%7D%5E%7B%5Cpi%2F2%7D%20%5Ccsc%20x%5C%2C%20dx%20%3D%20%5B%5Cln%7C%5Ccsc%20x%20-%20%5Ccot%20x%7C%5D%7B%5Cpi%2F6%7D%5E%7B%5Cpi%2F2%7D%20%3D%20%5Cln%282%2B%5Csqrt%7B3%7D%29).

Prob. 372. $sol091221 - 图56$ %7D%7Bx%7Ddx%20%3D%20%5Cfrac%7B3%7D%7B2%7D%5Cln%20(10)#card=math&code=%5Cint%7B10%7D%5E%7B100%7D%20%5Cfrac%7B%5Clog%7B10%7D%28x%29%7D%7Bx%7Ddx%20%3D%20%5Cfrac%7B3%7D%7B2%7D%5Cln%20%2810%29). Here we use $sol091221 - 图57$ %20%3D%20%5Cln%20(x)%2F%5Cln%20(10)#card=math&code=%5Clog_%7B10%7D%28x%29%20%3D%20%5Cln%20%28x%29%2F%5Cln%20%2810%29).