# Math Help - Integration by parts problem

1. ## 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

2. 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

3. 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 ?

4. 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 $(4x+4)^{13} = [4(x+1)]^{13} = 4^{13}(x+1)^{13}$

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

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

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

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

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

5. Excellent, thanks

6. 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:

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

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

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

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

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

7. Thanks

8. 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:

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

therefore your integral becomes:

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

9. Thanks again people ..