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# integration by parts watch

1. 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
2. I did it from scratch as it's clearer.
Edit: Obviously the last line factorises to the required expression.
Attached Images

3. arr...i see

if in the exam I had just figured ou the missing lin would I have still got the marks?!
4. 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!

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.

5. (Original post by 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!

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.

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
6. (Original post by 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!

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.

You have a function and a scalar multiple of derivative.
Formally:
Attached Images

7. (Original post by Gaz031)
You have a function and a scalar multiple of derivative.
Formally:
good point, I think guessing is more fun though!
8. Thanks so much...

It's mucho clearer now!

Back to revising...

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