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integration by parts

integration by parts

hi, this is from P5 aqa june 03. I can do most of it its just the penultimate line

Question..show

int.(tlnt)dt = 1/4.t^2(2lnt - 1) + c

so I did

int.(tlnt)dt

lnt.t^2/2 - int.t^2/2.1/x

....missing line

1/4t^2 (2lnt - 1)

I can see therefore that the missing line is

lnt.t^2/2 - 1/4t^2

can someone explain how this was integrated??

thanks
Reply 1
Avatar for Gaz031
Gaz031
15
I did it from scratch as it's clearer.
Edit: Obviously the last line factorises to the required expression.
IBP.gif3.2KB
Reply 2
Avatar for tom_austin
tom_austin
OP
arr...i see

if in the exam I had just figured ou the missing lin would I have still got the marks?!
Reply 3
Avatar for fabz
fabz
You probably would have! They don't know that you did it backwards, I always have to do that for proving stuff but they can't penalise you! :smile:

Please could you help me with this integrating trig???

∫sin² 2x. cos 2x dx

The answer is 1/6 sin³ 2x + c but I have no idea how to do it.

Thanks in advance! :biggrin:
Reply 4
Avatar for Master Gee
Master Gee
fabz
You probably would have! They don't know that you did it backwards, I always have to do that for proving stuff but they can't penalise you! :smile:

Please could you help me with this integrating trig???

∫sin² 2x. cos 2x dx

The answer is 1/6 sin³ 2x + c but I have no idea how to do it.

Thanks in advance! :biggrin:


Yo

With stuff like that you basically have to guess what do I differentiate to get that. If you ever have a trig function to a power (sin² 2x) with its differential to the power of one (cosx) next to it you can guess its integral:

Just take the trig bit with the power, add one to it and differentiate it and see what happens -

differentiate (sin2x)^3 = [(sin2x)^2] x 3 x 2cos2x

that's x3 from the ^3 and x 2cos2x from differentiating sin2x.

This gives you 6cosx(sin2x)^2 but you want cosx(sin2x)^2 so just divide the original bit by 6 to get:
1/6 sin³ 2x + C

Hope this helps
Reply 5
Avatar for Gaz031
Gaz031
15
fabz
You probably would have! They don't know that you did it backwards, I always have to do that for proving stuff but they can't penalise you! :smile:

Please could you help me with this integrating trig???

∫sin² 2x. cos 2x dx

The answer is 1/6 sin³ 2x + c but I have no idea how to do it.

Thanks in advance! :biggrin:


You have a function and a scalar multiple of derivative.
Formally:
int.gif3.1KB
Reply 6
Avatar for Master Gee
Master Gee
Gaz031
You have a function and a scalar multiple of derivative.
Formally:


good point, I think guessing is more fun though!
Reply 7
Avatar for fabz
fabz
Thanks so much...

It's mucho clearer now! :smile:

Back to revising... :frown:

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