# Integration by Parts and Arclength

• Feb 3rd 2010, 02:19 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. $\int^{e^4}_e\frac{1}{xln(x)}$ dx

My Solution

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

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

Problem 2. $\int$xsin4x dx

My Solution

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

$x(\frac{-1}{4}cos4x) - \int\frac{-1}{4}cos4x$ dx = $\frac{-x}{4}cos4x + \frac{1}{4}\int$ $cos4x$ dx = $\frac{-x}{4}cos4x + \frac{1}{4}(\frac{1}{4}sin4x) + C$ = $\frac{-x}{4} + \frac{1}{16}sin4x + C$ = $\frac{-4x + sin4x}{16} + C$

Problem 3. $\int\frac{7x}{4 + 9x^2}$ dx

My Solution

u = 4 + 9 $x^2$, du = 18x dx, dx = \frac{1}{18}x du

$\int\frac{7x}{u}\frac{1}{18x}$ du = $\int\frac{7}{18u}$ du = $\frac{7}{18}\int\frac{1}{u}$ du = $\frac{7}{18}ln|u| + C$ = $\frac{7}{18}ln(4 + 9x^2) + C$

Problem 4. $\int\frac{5}{3x^2 - 12x + 24}$ dx

My Solution

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

Problem 5. Calculate the length along the curve x = $cos^3(t)$, y = $sin^3(t)$, $0 \leq t \leq \frac{\pi}{2}$

$\frac{dx}{dt}$ = $-cos^2(t)sin(t)$, $\frac{dy}{dt}$ = $sin^2(t)cos(t)$

L = $\int^{\frac{\pi}{2}}_0\sqrt{(-cos^2(t)sin(t))^2 + (sin^2(t)cos(t))^2}$ dt = $\int^{\frac{\pi}{2}}_0\sqrt{cos^4(t)sin^2(t) + sin^4(t)cos^2(t)}$ dt = $\int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)(cos^2 (t) + sin^2(t)}$ dt = $\int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)}$ dt = $\int^{\frac{\pi}{2}}_0$cos(t)sin(t) dt

u = cos(t), du = -sin(t) dt, dt = $\frac{-1}{sin(t)}$ du

$-\int$ $usin(t)\frac{1}{sin(t)}$ du = $-\int$u du = $\frac{-u^2}{2} + C$ = $-[\frac{cos^2(t)}{2}]^{\frac{\pi}{2}}_0$ = -[0 - $\frac{1}{2}$] = $\frac{1}{2}$

Once again, thank you very much in advance.
• Feb 3rd 2010, 03:08 PM
Calculus26
Good Work

only thing:

1) ln(ln(e^4)) = ln(4)
• Feb 3rd 2010, 03:50 PM
skeeter
Quote:

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

My Solution

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

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

...
• Feb 3rd 2010, 10:12 PM
mturner07
Thank you. My new answer is $\frac{5}{6}arctan(\frac{x-2}{2}) + C$ Is this correct?
• Feb 3rd 2010, 10:58 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:

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

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, 05:20 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. $\int^{e^4}_e\frac{1}{xln(x)}$ dx

My Solution

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

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

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 $x= e^4$, [tex]u= ln(e^4)= 4. The integral becomes $\int_1^4 \frac{1}{u}du$

Quote:

Problem 2. $\int$xsin4x dx

My Solution

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

$x(\frac{-1}{4}cos4x) - \int\frac{-1}{4}cos4x$ dx = $\frac{-x}{4}cos4x + \frac{1}{4}\int$ $cos4x$ dx = $\frac{-x}{4}cos4x + \frac{1}{4}(\frac{1}{4}sin4x) + C$ = $\frac{-x}{4} + \frac{1}{16}sin4x + C$ = $\frac{-4x + sin4x}{16} + C$

Problem 3. $\int\frac{7x}{4 + 9x^2}$ dx

My Solution

u = 4 + 9 $x^2$, du = 18x dx, dx = \frac{1}{18}x du

$\int\frac{7x}{u}\frac{1}{18x}$ du = $\int\frac{7}{18u}$ du = $\frac{7}{18}\int\frac{1}{u}$ du = $\frac{7}{18}ln|u| + C$ = $\frac{7}{18}ln(4 + 9x^2) + C$

Problem 4. $\int\frac{5}{3x^2 - 12x + 24}$ dx

My Solution

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

Problem 5. Calculate the length along the curve x = $cos^3(t)$, y = $sin^3(t)$, $0 \leq t \leq \frac{\pi}{2}$

$\frac{dx}{dt}$ = $-cos^2(t)sin(t)$, $\frac{dy}{dt}$ = $sin^2(t)cos(t)$
L = $\int^{\frac{\pi}{2}}_0\sqrt{(-cos^2(t)sin(t))^2 + (sin^2(t)cos(t))^2}$ dt = $\int^{\frac{\pi}{2}}_0\sqrt{cos^4(t)sin^2(t) + sin^4(t)cos^2(t)}$ dt = $\int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)(cos^2 (t) + sin^2(t)}$ dt = $\int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)}$ dt = $\int^{\frac{\pi}{2}}_0$cos(t)sin(t) dt
u = cos(t), du = -sin(t) dt, dt = $\frac{-1}{sin(t)}$ du
$-\int$ $usin(t)\frac{1}{sin(t)}$ du = $-\int$u du = $\frac{-u^2}{2} + C$ = $-[\frac{cos^2(t)}{2}]^{\frac{\pi}{2}}_0$ = -[0 - $\frac{1}{2}$] = $\frac{1}{2}$