Thread: Integration by parts problem

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

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

Thanks

Originally Posted by wolfhound

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

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

Thanks

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

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

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

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


edit: please check my signs because I'm tend to get an embarrassing number of them wrong

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 ?

You could always use the binomial theorem to expand

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

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

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

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

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

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


$\displaystyle 4^{13}$ is a massive number but it is, and thus acts like, a constant

Excellent, thanks

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:

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

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

integrate by parts:

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

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

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

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

$\displaystyle =-6x(4x+4)^{14}+ \dfrac{(4x+4)^{15}}{10}+C$

Thanks

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

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

use substution rule:

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

therefore your integral becomes:

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

Thanks again people ..