# Integration by parts problem

• December 2nd 2010, 10:50 AM
wolfhound
Integration by parts problem
Hi, Can someone tell me how to do this integration problem by parts please, its driving me mental .

$integrate 4tsin(4t-2)dt$

Thanks
• December 2nd 2010, 10:56 AM
e^(i*pi)
Quote:

Originally Posted by wolfhound
Hi, Can someone tell me how to do this integration problem by parts please, its driving me mental .

$integrate 4tsin(4t-2)dt$

Thanks

$u = 4t \implies u' = 4$

$v' = \sin(4t-2) \implies v = -0.25\cos(4t-2)$

$\int udv = uv - \int vdu = -t\cos(4t-2) + \int \cos(4t-2)$

The integral of $\cos(4t-2)$ is a standard trig integral

edit: please check my signs because I'm tend to get an embarrassing number of them wrong
• December 2nd 2010, 11:07 AM
wolfhound
Thank you very much for that!
while we are on the subject, one more problem please

This integrate : -336x(4x+4)^(13) dx

u=-336x du=-336
dv=(4x+4)^(13)
v= (4x+4)^(14)
...... 14

and I got -24x(4x+4)^(14) +24(4x + 4)^(15) + C
.... ................ .................... 60

I think my answer is wrong ?
• December 2nd 2010, 11:25 AM
e^(i*pi)
You could always use the binomial theorem to expand (Wink)

It looks like you haven't divided by 4 when finding v.

I would be inclined to say that $(4x+4)^{13} = [4(x+1)]^{13} = 4^{13}(x+1)^{13}$

$\int -336 \cdot 4^{13} x (x+1)^{13}$

$u = -336 \cdot 4^{13}x \implies du = -336 \cdot 4^{13}$

$dv = (x+1)^{13} \implies v = \dfrac{1}{14}(x+1)^{14}$

$-24\cdot 4^{13}x(x+1)^{14} + \int 24 \cdot 4^{13} (x+1)^{14}$

$4^{13}$ is a massive number but it is, and thus acts like, a constant
• December 2nd 2010, 11:31 AM
wolfhound
Excellent, thanks :)
• December 2nd 2010, 11:39 AM
harish21
Quote:

Originally Posted by wolfhound
Thank you very much for that!
while we are on the subject, one more problem please

This integrate : -336x(4x+4)^(13) dx

u=-336x du=-336
dv=(4x+4)^(13)
v= (4x+4)^(14)
...... 14

and I got -24x(4x+4)^(14) +24(4x + 4)^(15) + C
.... ................ .................... 60

I think my answer is wrong ?

Looks like you made calculation errors:

$Let \; u = -336x \;\;\;\;\;du=-336\;dx$

$dv = (4x+4)^{13}\;dx \;\;\;\;\; v = \dfrac{(4x+4)^{14}}{56}$

integrate by parts:

$\displaystyle \int u\;dv= uv - \int v\;du$

$= \displaystyle -336x \cdot \dfrac{(4x+4)^{14}}{56} \;-\;\int \dfrac{(4x+4)^{14}}{56}\;(-336)\;dx$

$=\displaystyle -6x(4x+4)^{14}+6 \int {(4x+4)^{14}} \;dx$

$=\displaystyle -6x(4x+4)^{14}+ 6 \cdot \dfrac{(4x+4)^{15}}{60}+C$

$=-6x(4x+4)^{14}+ \dfrac{(4x+4)^{15}}{10}+C$
• December 2nd 2010, 11:44 AM
wolfhound
Thanks :)
• December 2nd 2010, 11:48 AM
harish21
Quote:

Originally Posted by wolfhound
Thanks, but where did 56 come from when you integrate dv should it not be that v= ((4x+4)^14)/14

$\displaystyle \int (4x+4)^{13}\;dx$

use substution rule:

$Let\;\; a = 4x+4 \; \rightarrow da = 4 dx \rightarrow dx = \dfrac{da}{4}$

$\displaystyle \int a^{13} \dfrac{da}{4} = \dfrac{1}{4} \int a^{13}\;da = \dfrac{1}{4} \times \dfrac{a^{14}}{14} = \dfrac{a^{14}}{56}=\dfrac{(4x+4)^{14}}{56}$