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

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- Dec 2nd 2010, 10:50 AMwolfhoundIntegration 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 - Dec 2nd 2010, 10:56 AMe^(i*pi)
$\displaystyle u = 4t \implies u' = 4$

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

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

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 - Dec 2nd 2010, 11:07 AMwolfhound
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 ? - Dec 2nd 2010, 11:25 AMe^(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 $\displaystyle (4x+4)^{13} = [4(x+1)]^{13} = 4^{13}(x+1)^{13}$

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

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

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

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

$\displaystyle 4^{13}$ is a massive number but it is, and thus acts like, a constant - Dec 2nd 2010, 11:31 AMwolfhound
Excellent, thanks :)

- Dec 2nd 2010, 11:39 AMharish21
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:

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

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

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

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

$\displaystyle =-6x(4x+4)^{14}+ \dfrac{(4x+4)^{15}}{10}+C$ - Dec 2nd 2010, 11:44 AMwolfhound
Thanks :)

- Dec 2nd 2010, 11:48 AMharish21
$\displaystyle \displaystyle \int (4x+4)^{13}\;dx$

use substution rule:

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

therefore your integral becomes:

$\displaystyle \displaystyle \int a^{13} \dfrac{da}{4} = \dfrac{1}{4} \int a^{13}\;da = \dfrac{1}{4} \times \dfrac{a^{14}}{14} = \dfrac{a^{14}}{56}=\dfrac{(4x+4)^{14}}{56}$ - Dec 2nd 2010, 11:55 AMwolfhound
Thanks again people:) ..