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# AEA Integral watch

1. After slogging my through the first two parts of the question in a whole hour , I have become stuck at part c. Its worth 9 marks and I just find a way to Integrate this.

I have tried the sub u = cos x, and got to Integrate (√(2u^2 - 1) du). ANd it certianly doesn't work by parts.

Thanks,
Attached Images

2. Is this part of the July 2005 AEA paper? If so, you can use the substitution u=sinx and then use earlier results to evaluate the integral.
3. okay, this is what i did, it may be wrong though... (i'm sure i've done somthing wrong to be honest but this is what i would have done)

sin2x ≡ 2sinxcosx so it follows that

sin2x/2cosx ≡ sinx

so rewrite the integral

∫ (sin2x/2cos2x) . √cos2x dx

then

∫(sin2x/2√cos2x)dx

then let u = √cos2x

so that du/dx = -sin2x/√cos2x which is actually -sin2x/u

then you have that du = (-sin2x/u)dx

and

-1/2du = (sin2x/2u) dx

going back to the original integral:

substitute u for √cos2x

∫(sinx/2u) dx

and thus

∫ -1/2 du

or -1/2∫1du

now to convert limits which are in x, u = √cos 2(π/4) = 0 and u = √cos 2(0) = 1

then you have an integral:

limits 0, 1

-1/2∫1du = -1/2{[u]}

= -1/2 {[1] - [0]}

= -1/2
4. I think you make a mistake from
sin(2x)/2cosx = sinx and inserting it into integral because u make
cosx - ->cos(2x).
5. im not really sure about this but u no u can rewrite cos 2x as cos^2(x)-1. so can u see what to do next?
6. so can u see what to do next?

No lol. Thats exactly what I did, to get root of (2u^2 - 1) du..

Gaz, how do you mean, use part of the question (earlier results)?
7. aha!, you're right, i knew that solution was too simple. thanks for pointing that out.
i am smacking my head right now... that is quite a big mistake (i can't believe i did that!!!)
8. (Original post by Vazzyb)
so can u see what to do next?

No lol. Thats exactly what I did, to get root of (2u^2 - 1) du..

Gaz, how do you mean, use part of the question (earlier results)?
lol sorry im wrong just ignore me...
9. (Original post by Vazzyb)
so can u see what to do next?

No lol. Thats exactly what I did, to get root of (2u^2 - 1) du..

Gaz, how do you mean, use part of the question (earlier results)?
I remembered doing this question in the exam and I'm fairly sure I used the earlier parts. I can't remember the specifics though so people might be able to help better if you posted the entire question.
10. OK cheers, The whole question...

[Attached]

(sorry about not doing it earlier!)
Attached Images

11. (Original post by Gaz031)
I remembered doing this question in the exam and I'm fairly sure I used the earlier parts. I can't remember the specifics though so people might be able to help better if you posted the entire question.
Well after gaz's clue i think you have to rewrite cos2x as 1- 2sin^2x
so ∫sinx√cos2x=∫sinx√(1- 2sin^2x)
Then i would be able to do it by recognition because im not very good with substitution... Do u guys know any resources that help with integration by substitution?

Edit:sorry im wrong again lol please do ignore me
12. Looking at it, I think there's a fairly straight forward plan of attack:

Where I've used y2=2u2. Rearranging gives:

13. (Original post by Wrangler)
Looking at it, I think there's a fairly straight forward plan of attack:

Where I've used y2=2u2. Rearranging gives:

can't be right, since this paper only requires c1-c4 knowledge.
14. (Original post by Widowmaker)
can't be right, since this paper only requires c1-c4 knowledge.
Well, I'm pretty sure that's right. It may not be the way you would have gone about it.
15. have u tried using the substitution u = √2cosx?
16. i think i got it:

cos2x = 2cos²x - 1

so ∫sinx√(2cos²x -1) dx

then let u = √2cosx

then du/dx = -√2sinx

then you get:

-√2∫√u²-1 du with limits when x = π/4 u = 1, and when x = 0, u = √2

then make substitution u = sec θ,

giving (from (a)) ∫secθtan²θdθ with limits, θ = π/4 and θ = 0

then if you did part (b), you just substitute in and you're done!

--------------

HEY!, that was my idea
17. (Original post by Mr. Pink)
i think i got it:

cos2x = 2cos²x - 1

so ∫sinx√(2cos²x -1) dx

then let u = √2cosx

then du/dx = -√2sinx

then you get:

-√2∫√u²-1 du with limits when x = π/4 u = 1, and when x = 0, u = √2

then make substitution u = sec θ,

giving (from (a)) ∫secθtan²θdθ with limits, θ = π/4 and θ = 0

then if you did part (b), you just substitute in and you're done!

--------------

HEY!, that was my idea
lol i guess it was, i was about to do it but i suck at substitution
18. i'm glad i figured it out though, that other mistake i made before was REALLY bad
19. (Original post by Mr. Pink)
yeah i made quite a few mistakes as well, i thought the derivative of sin^2x was 2 sinx lol. I need more sleep....
20. edit : nevermind eh?

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