Integration by Parts and Arclength

• Feb 3rd 2010, 01:19 PM
mturner07
Integration by Parts and Arclength
Back again! This time I have five problems that I have completed that, if anyone would be so kind, I would like to have looked over and corrected if they are wrong. Thank you very very much in advance!

Problem 1. $\displaystyle \int^{e^4}_e\frac{1}{xln(x)}$ dx

My Solution

u = ln(x), du = $\displaystyle \frac{1}{x}$ dx, dx = x du

$\displaystyle \int\frac{1}{xu}x$ du = $\displaystyle \int\frac{1}{u}$ du = ln(u) + C = $\displaystyle [ln(lnx)]^{e_4}_e$ = ln[ln($\displaystyle e^4$)] - ln[ln(e)] = ln[ln($\displaystyle e^4$)] - 0 = ln[ln($\displaystyle e^4$)]

Problem 2. $\displaystyle \int$xsin4x dx

My Solution

u = x, du = dx, dv = sin4x, v = $\displaystyle \frac{-1}{4}$cos4x

$\displaystyle x(\frac{-1}{4}cos4x) - \int\frac{-1}{4}cos4x$ dx = $\displaystyle \frac{-x}{4}cos4x + \frac{1}{4}\int$$\displaystyle cos4x dx = \displaystyle \frac{-x}{4}cos4x + \frac{1}{4}(\frac{1}{4}sin4x) + C = \displaystyle \frac{-x}{4} + \frac{1}{16}sin4x + C = \displaystyle \frac{-4x + sin4x}{16} + C Problem 3. \displaystyle \int\frac{7x}{4 + 9x^2} dx My Solution u = 4 + 9\displaystyle x^2, du = 18x dx, dx = \frac{1}{18}x du \displaystyle \int\frac{7x}{u}\frac{1}{18x} du = \displaystyle \int\frac{7}{18u} du = \displaystyle \frac{7}{18}\int\frac{1}{u} du = \displaystyle \frac{7}{18}ln|u| + C = \displaystyle \frac{7}{18}ln(4 + 9x^2) + C Problem 4. \displaystyle \int\frac{5}{3x^2 - 12x + 24} dx My Solution \displaystyle \frac{1}{3}\int\frac{5}{x^2 - 4x + 8} dx = \displaystyle \frac{5}{3}\int\frac{1}{x^2 - 4x + 8} dx = \displaystyle \frac{5}{3}\int\frac{1}{(x + 4)^2 + 4} dx = \displaystyle \frac{5}{3}[\frac{1}{2}arctan(\frac{x + 4}{2})] + C = \displaystyle \frac{5}{6}arctan(\frac{x + 4}{2}) + C Problem 5. Calculate the length along the curve x = \displaystyle cos^3(t), y = \displaystyle sin^3(t), \displaystyle 0 \leq t \leq \frac{\pi}{2} My Solution (Not so sure about this one.) \displaystyle \frac{dx}{dt} = \displaystyle -cos^2(t)sin(t), \displaystyle \frac{dy}{dt} = \displaystyle sin^2(t)cos(t) L = \displaystyle \int^{\frac{\pi}{2}}_0\sqrt{(-cos^2(t)sin(t))^2 + (sin^2(t)cos(t))^2} dt = \displaystyle \int^{\frac{\pi}{2}}_0\sqrt{cos^4(t)sin^2(t) + sin^4(t)cos^2(t)} dt = \displaystyle \int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)(cos^2 (t) + sin^2(t)} dt = \displaystyle \int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)} dt = \displaystyle \int^{\frac{\pi}{2}}_0cos(t)sin(t) dt u = cos(t), du = -sin(t) dt, dt = \displaystyle \frac{-1}{sin(t)} du \displaystyle -\int$$\displaystyle usin(t)\frac{1}{sin(t)}$ du = $\displaystyle -\int$u du = $\displaystyle \frac{-u^2}{2} + C$ = $\displaystyle -[\frac{cos^2(t)}{2}]^{\frac{\pi}{2}}_0$ = -[0 - $\displaystyle \frac{1}{2}$] = $\displaystyle \frac{1}{2}$

Once again, thank you very much in advance.
• Feb 3rd 2010, 02:08 PM
Calculus26
Good Work

only thing:

1) ln(ln(e^4)) = ln(4)
• Feb 3rd 2010, 02:50 PM
skeeter
Quote:

Originally Posted by mturner07
Problem 4. $\displaystyle \int\frac{5}{3x^2 - 12x + 24}$ dx

My Solution

$\displaystyle \frac{1}{3}\int\frac{5}{x^2 - 4x + 8}$ dx = $\displaystyle \frac{5}{3}\int\frac{1}{x^2 - 4x + 8}$ dx = $\displaystyle \frac{5}{3}\int\frac{1}{(x + 4)^2 + 4}$ dx = $\displaystyle \frac{5}{3}[\frac{1}{2}arctan(\frac{x + 4}{2})] + C$ = $\displaystyle \frac{5}{6}arctan(\frac{x + 4}{2}) + C$

$\displaystyle \textcolor{red}{\frac{1}{x^2-4x+8} = \frac{1}{(x-2)^2 + 4}}$

...
• Feb 3rd 2010, 09:12 PM
mturner07
Thank you. My new answer is $\displaystyle \frac{5}{6}arctan(\frac{x-2}{2}) + C$ Is this correct?
• Feb 3rd 2010, 09:58 PM
buckeye1973
In the first problem notice that the limits of integration can also go through the u-substitution process, and you don't need to return to terms of x. Since e and e^4 are in terms of x, and u = ln(x), then:

$\displaystyle \int_e^{e^4}\frac{1}{xln(x)} dx = \int_{ln(e)}^{ln(e^4)}\frac{1}{u} du = \int_{1}^{4}\frac{1}{u} du = ln(u)_{1}^{4} = ln(4) - ln(1) = ln(4)$

It isn't always easy or better to convert your limits of integration to the terms of your substitution, but sometimes it does simplify things.

Brian
(Who is only about a semester past this point, so I hope what I've said is all correct!)
• Feb 4th 2010, 04:20 AM
HallsofIvy
Quote:

Originally Posted by mturner07
Back again! This time I have five problems that I have completed that, if anyone would be so kind, I would like to have looked over and corrected if they are wrong. Thank you very very much in advance!

Problem 1. $\displaystyle \int^{e^4}_e\frac{1}{xln(x)}$ dx

My Solution

u = ln(x), du = $\displaystyle \frac{1}{x}$ dx, dx = x du

$\displaystyle \int\frac{1}{xu}x$ du = $\displaystyle \int\frac{1}{u}$ du = ln(u) + C = $\displaystyle [ln(lnx)]^{e_4}_e$ = ln[ln($\displaystyle e^4$)] - ln[ln(e)] = ln[ln($\displaystyle e^4$)] - 0 = ln[ln($\displaystyle e^4$)]

It is far simpler just to use substitution. Let u= ln(x) so that du= dx/x. When x= e, u= ln(e)= 1. When $\displaystyle x= e^4$, [tex]u= ln(e^4)= 4. The integral becomes $\displaystyle \int_1^4 \frac{1}{u}du$

Quote:

Problem 2. $\displaystyle \int$xsin4x dx

My Solution

u = x, du = dx, dv = sin4x, v = $\displaystyle \frac{-1}{4}$cos4x

$\displaystyle x(\frac{-1}{4}cos4x) - \int\frac{-1}{4}cos4x$ dx = $\displaystyle \frac{-x}{4}cos4x + \frac{1}{4}\int$$\displaystyle cos4x dx = \displaystyle \frac{-x}{4}cos4x + \frac{1}{4}(\frac{1}{4}sin4x) + C = \displaystyle \frac{-x}{4} + \frac{1}{16}sin4x + C = \displaystyle \frac{-4x + sin4x}{16} + C Problem 3. \displaystyle \int\frac{7x}{4 + 9x^2} dx My Solution u = 4 + 9\displaystyle x^2, du = 18x dx, dx = \frac{1}{18}x du \displaystyle \int\frac{7x}{u}\frac{1}{18x} du = \displaystyle \int\frac{7}{18u} du = \displaystyle \frac{7}{18}\int\frac{1}{u} du = \displaystyle \frac{7}{18}ln|u| + C = \displaystyle \frac{7}{18}ln(4 + 9x^2) + C Problem 4. \displaystyle \int\frac{5}{3x^2 - 12x + 24} dx My Solution \displaystyle \frac{1}{3}\int\frac{5}{x^2 - 4x + 8} dx = \displaystyle \frac{5}{3}\int\frac{1}{x^2 - 4x + 8} dx = \displaystyle \frac{5}{3}\int\frac{1}{(x + 4)^2 + 4} dx = \displaystyle \frac{5}{3}[\frac{1}{2}arctan(\frac{x + 4}{2})] + C = \displaystyle \frac{5}{6}arctan(\frac{x + 4}{2}) + C Problem 5. Calculate the length along the curve x = \displaystyle cos^3(t), y = \displaystyle sin^3(t), \displaystyle 0 \leq t \leq \frac{\pi}{2} My Solution (Not so sure about this one.) \displaystyle \frac{dx}{dt} = \displaystyle -cos^2(t)sin(t), \displaystyle \frac{dy}{dt} = \displaystyle sin^2(t)cos(t) L = \displaystyle \int^{\frac{\pi}{2}}_0\sqrt{(-cos^2(t)sin(t))^2 + (sin^2(t)cos(t))^2} dt = \displaystyle \int^{\frac{\pi}{2}}_0\sqrt{cos^4(t)sin^2(t) + sin^4(t)cos^2(t)} dt = \displaystyle \int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)(cos^2 (t) + sin^2(t)} dt = \displaystyle \int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)} dt = \displaystyle \int^{\frac{\pi}{2}}_0cos(t)sin(t) dt u = cos(t), du = -sin(t) dt, dt = \displaystyle \frac{-1}{sin(t)} du \displaystyle -\int$$\displaystyle usin(t)\frac{1}{sin(t)}$ du = $\displaystyle -\int$u du = $\displaystyle \frac{-u^2}{2} + C$ = $\displaystyle -[\frac{cos^2(t)}{2}]^{\frac{\pi}{2}}_0$ = -[0 - $\displaystyle \frac{1}{2}$] = $\displaystyle \frac{1}{2}$

Once again, thank you very much in advance.