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# Test Ourselves: P3 watch

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1. With the exam on wednesday, i thought i'd start this thread for us test each other with questions to see how ready we are! I'll start the ball rolling with a question im having trouble with...

prove that the integral of xcos2x can be expressed as 1/2(sinx(2xcosx-sinx) + c

anyone...?
2. (Original post by scottyoneill)
With the exam on wednesday, i thought i'd start this thread for us test each other with questions to see how ready we are! I'll start the ball rolling with a question im having trouble with...

prove that the integral of xcos2x can be expressed as 1/2(sinx(2xcosx-sinx) + c

anyone...?
dy/dx= 1 and ∫cos2x =sin2x/2

xsin2x/ 2 - ∫ sin2x/2

xsin2x/2 - (- cos 2x /4)

xsin2x/2 + cos2x /4

sin2A=2sinAcosA and cos2x =1 -sin^2x

2xsinxcosx + 1 - sin^2x /4

then im stuck where did i go wrong
3. Just use integration by parts with u=x and v´=Cos 2x
ie u´=1 and v = 1/2 Sin 2x

therefore

I xCos2x dx = x . 1/2 Sin 2x - I I/2 Sin 2x dx

= 1/2 x 2 Sin x . Cos x + 1/4 Cos 2x + c

= 1/2 x 2 Sin x.Cos x + 1/4 (1 - 2 Sin^2 x) +c

= 1/2 x 2 Sin x.Cos x + 1/4 - 1/2 Sin^2 x +c

= 1/2 x 2 Sin x.Cos x - 1/2 Sin^2 x + k

= 1/2(sinx(2xcosx-sinx) + k
4. (Original post by scottyoneill)
With the exam on wednesday, i thought i'd start this thread for us test each other with questions to see how ready we are! I'll start the ball rolling with a question im having trouble with...

prove that the integral of xcos2x can be expressed as 1/2(sinx(2xcosx-sinx) + c

anyone...?
∫xcos2x dx

Using integration by parts I got:

(xsin2x)/2 + (cos2x)/4

Cross multiplied everything to obtain:

(4xsin2x+2cos2x)/8

then expanded the sin2x and cos2x to get:

(2xsinxcosx)/2 - (1-2sin^2x)/4

[8xsinxcosx-(2-4sin^2x)]/8

(2xsinxcosx-sin^2x)/2
=[sinx(2xcosx-sinx)]/2
5. here's a quickie

∫ √(1-x^2) dx

.. actually this may be more complicated than I thought...
6. let x = sin u etc
7. (Original post by Mrm.)
let x = sin u etc
The is the answer just x??? Seems very unlikely - but that's what i got!!!
8. (Original post by kimoni)
here's a quickie

∫ √(1-x^2) dx

.. actually this may be more complicated than I thought...
-3x(1-x^2)^3/2 + c
9. (Original post by Mrm.)
let x = sin u etc
yeah, that's what I thought, and I get an answer, but go to http://www.calc101.com/webMathematica/integrals.jsp and type in sqrt[1-x^2] you get something completely different.. or is it just me?

ed: actually.. maybe not completely different
10. (Original post by Silly Sally)
The is the answer just x??? Seems very unlikely - but that's what i got!!!
Ok then!!! Got it completely wrong!!! can someone show me how to do this question please!!!
11. i dont have a password
12. (Original post by Mrm.)
neither do I, but it still shows you the answer

-can anyone offer a definitive proof for this question?
13. On the Edexcel board its it always presumed that x=sinu?
14. i get something very similar to the calc101 answer....just sorting out the final bits and pieces

you do use the sub x=sin u
15. I just did it and it comes out fine.
16. (Original post by Bhaal85)
∫xcos2x dx

Using integration by parts I got:

(xsin2x)/2 + (cos2x)/4

Cross multiplied everything to obtain:

(4xsin2x+2cos2x)/8

then expanded the sin2x and cos2x to get:

(2xsinxcosx)/2 - (1-2sin^2x)/4

[8xsinxcosx-(2-4sin^2x)]/8

(2xsinxcosx-sin^2x)/2
=[sinx(2xcosx-sinx)]/2
Sorry i don't understand - what you did for the bold bit? Can you please explain
17. Well having used x=sin u as my sub, i get

∫cos^2 x
= (sin2u)/4 + u/2

x=sin u
u=arcsin x

sin2(sin^-1 x)/4 + (sin^-1)/2 + C

which is almost there.. I dont know how to do sin2(sin^-1 x)/4 though
18. (Original post by Mrm.)
I just did it and it comes out fine.
can you show your working then?
19. (Original post by kimoni)
Well having used x=sin u as my sub, i get

∫cos^2 x
= (sin2u)/4 + u/2

x=sin u
u=arcsin x

sin2(sin^-1 x)/4 + (sin^-1)/2 + C

which is almost there.. I dont know how to do sin2(sin^-1 x)/4 though
Whats happened to the square root part when you integrate?
20. (Original post by Silly Sally)
Whats happened to the square root part when you integrate?
If you use x = sin t in (1 - x^2)^1/2, you get cos t (from the identity sin^2 t + cos^2 t = 1).

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