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Core 4 Soloman Paper D

Hi.
For question 1 what integration method should i be using.
https://5c59854d0ccd29d489c9e5e689a8bbadf49aa0f0.googledrive.com/host/0B1ZiqBksUHNYREhxMHhfam1IQm8/for-OCR/Solomon%20D%20QP%20-%20C4%20OCR.pdf
thanks
Reply 1
Avatar for Zacken
Zacken
22
Original post by SamuelN98


No set one, there's a variety of methods. You could expand out and say ∫sin⁡x+sin⁡xcos⁡x dx=−cos⁡x+sin⁡2x2+C\int \sin x + \sin x \cos x \, \mathrm{d}x = -\cos x + \frac{\sin^2 x}{2} + \mathcal{C} by inspection or ∫sin⁡x+sin⁡xcos⁡x dx=−cos⁡x−cos⁡2x2+C\int \sin x + \sin x\cos x \, \mathrm{d}x = -\cos x - \frac{\cos^2 x}{2} + \mathcal{C} by inspection or note that sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \frac{\sin 2x}{2}, or use the substitution u=cos⁡xu = \cos x on the latter half of the integral, or u=sin⁡xu = \sin x, etc...
(edited 8 years ago)
Reply 2
Avatar for Notnek
Notnek
21
Original post by SamuelN98

Expand and use the identity:

sin⁡2A≡2sin⁡Acos⁡A\displaystyle \sin 2A \equiv 2\sin A \cos A
Reply 3
Avatar for SamuelN98
SamuelN98
OP
5
Original post by Zacken
No set one, there's a variety of methods. You could expand out and say ∫sin⁡x+sin⁡xcos⁡x dx=−cos⁡x+sin⁡2x2+C\int \sin x + \sin x \cos x \, \mathrm{d}x = -\cos x + \frac{\sin^2 x}{2} + \mathcal{C} by inspection or ∫sin⁡x+sin⁡xcos⁡x dx=−cos⁡x−cos⁡2x2+C\int \sin x + \sin x\cos x \, \mathrm{d}x = -\cos x - \frac{\cos^2 x}{2} + \mathcal{C} by inspection or note that sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \frac{\sin 2x}{2}, or use the substitution u=cos⁡xu = \cos x on the latter half of the integral, or u=sin⁡xu = \sin x, etc...


Sorry , could how explain how you got (cos^2)(x)/2 ,is that the reverse chain rule?
Reply 4
Avatar for Zacken
Zacken
22
Original post by SamuelN98
Sorry , could how explain how you got (cos^2)(x)/2 ,is that the reverse chain rule?


Yeah.
Reply 5
Avatar for SamuelN98
SamuelN98
OP
5
Original post by Zacken
Yeah.


Hey.
Would you be able to help me understand what is meant in question 8.
I cant visualize what is mean by point c lies on l2 such that AC is perpendicular to BC.
Reply 6
Avatar for Zacken
Zacken
22
Original post by SamuelN98
Hey.
Would you be able to help me understand what is meant in question 8.
I cant visualize what is mean by point c lies on l2 such that AC is perpendicular to BC.


Crap drawing:


Green is line 1, blue is line 2.
@Zacken Is it possible to use integration by parts with this?
Reply 8
Avatar for SamuelN98
SamuelN98
OP
5
Original post by Zacken
Crap drawing:


Green is line 1, blue is line 2.

Thanks.
Reply 9
Avatar for Zacken
Zacken
22
Original post by Don Pedro K.
@Zacken Is it possible to use integration by parts with this?


Yep, that's definitely another method.
Original post by Zacken
Yep, that's definitely another method.


Ah okay thank you!

Oh yeah this is a bit random but, assuming that you did (which I think you did), how you find last year's C3 paper? I've heard complaints about it but was it really that bad?
Reply 11
Avatar for Zacken
Zacken
22
Original post by Don Pedro K.
Ah okay thank you!

Oh yeah this is a bit random but, assuming that you did (which I think you did), how you find last year's C3 paper? I've heard complaints about it but was it really that bad?


No worries.

Seemed fairly standard to me. :dontknow:
Original post by Zacken
No worries.

Seemed fairly standard to me. :dontknow:


Would you say that the papers generally get harder (even if only a bit) from the earlier ones to the later ones?
Reply 13
Avatar for Zacken
Zacken
22
Original post by Don Pedro K.
Would you say that the papers generally get harder (even if only a bit) from the earlier ones to the later ones?


Erm, I wouldn't really say so, no. They seem to fluctuate in difficulty, some of the recent ones area easier, most are standard, some are harder.
Original post by Zacken
Erm, I wouldn't really say so, no. They seem to fluctuate in difficulty, some of the recent ones area easier, most are standard, some are harder.


Ah okay awesome, thanks Zacken! Best of luck with things.
Reply 15
Avatar for Zacken
Zacken
22
Original post by Don Pedro K.
Ah okay awesome, thanks Zacken! Best of luck with things.


No problem, good luck to you too!

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