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# Test Ourselves: P3 watch

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1. right here it is....it's long and confusing and can be done better...but it works:

x=sinu
therefore 1-x^2 = cos^2 u
dx/du = cos u
dx = ducos u

therefore INT sqrt(1-x^2) = INT cos^2 u

integrate using cos^2u = .5(cos 2u +1)

and you get:

1/4 * sin 2u + 1/2 * u + c

1/4 * sin(2arcsinx) + 1/2 * arcsin x + c

but sin(2arcsinx) = 2sin(arcsinx)cos(arcsinx) (double ange formula)
but now we simplify cos(arcsin x)
substitue for x = sin u

cos(arcsin sin u) = cos(u)
cos^2(u) = 1-sin^2u
so cos(u) = sqrt(1-sin^2 u)
but sin u = x
so
cos(u) = cos(arcsin x) = sqrt(1-x^2)

so:
1/4 * sqrt(1-x^2) + 1/2 * arcsin x + c
2. (Original post by Silly Sally)
Sorry i don't understand - what you did for the bold bit? Can you please explain
(4xsin2x+2cos2x)/8

4x(sin2x)= 4x(2sinxcosx) = 8xsinxcosx

2(cos2x)=2(Cos^2x-Sin^2x) = 2((1-sin^2x)-sin^2x) = 2(1-2sin^2x) therefore:

(4xsin2x+2cos2x)/8 = (8xsinxcosx - 4sin^2x + 2)/8
3. www.xs4all.nl/~mcgee/int.jpg

that link should work.
MrM;
4. www.xs4all.nl/~mcgee/Int.jpg

that even
5. what happens after u have substituted 1-sin^2 u for cos^2 u ....where does the square root go?
6. (Original post by scottyoneill)
what happens after u have substituted 1-sin^2 u for cos^2 u ....where does the square root go?
Taken out as a factor.
7. (Original post by Willa)
right here it is....it's long and confusing and can be done better...but it works:

x=sinu
therefore 1-x^2 = cos^2 u
dx/du = cos u
dx = ducos u

therefore INT sqrt(1-x^2) = INT cos^2 u

integrate using cos^2u = .5(cos 2u +1)

and you get:

1/4 * sin 2u + 1/2 * u + c

1/4 * sin(2arcsinx) + 1/2 * arcsin x + c

but sin(2arcsinx) = 2sin(arcsinx)cos(arcsinx) (double ange formula)
but now we simplify cos(arcsin x)
substitue for x = sin u

cos(arcsin sin u) = cos(u)
cos^2(u) = 1-sin^2u
so cos(u) = sqrt(1-sin^2 u)
but sin u = x
so
cos(u) = cos(arcsin x) = sqrt(1-x^2)

so:
1/4 * sqrt(1-x^2) + 1/2 * arcsin x + c
excellent! cheers!
8. (Original post by Willa)
right here it is....it's long and confusing and can be done better...but it works:

x=sinu
therefore 1-x^2 = cos^2 u
dx/du = cos u
dx = ducos u

therefore INT sqrt(1-x^2) = INT cos^2 u

integrate using cos^2u = .5(cos 2u +1)

and you get:

1/4 * sin 2u + 1/2 * u + c

1/4 * sin(2arcsinx) + 1/2 * arcsin x + c

but sin(2arcsinx) = 2sin(arcsinx)cos(arcsinx) (double ange formula)
but now we simplify cos(arcsin x)
substitue for x = sin u

cos(arcsin sin u) = cos(u)
cos^2(u) = 1-sin^2u
so cos(u) = sqrt(1-sin^2 u)
but sin u = x
so
cos(u) = cos(arcsin x) = sqrt(1-x^2)

so:
1/4 * sqrt(1-x^2) + 1/2 * arcsin x + c
I am sure integrating arcsinx/getting arcsin x from an integral isn't on the edexcel sylabus for p3
9. #jeepers creepers, where did you get those eyes from, I said jeepers creepers...#
10. (Original post by Silly Sally)
I am sure integrating arcsinx/getting arcsin x from an integral isn't on the edexcel sylabus for p3
no matter
11. Some slightly more challenging ones:

ok, differentiate the following: x^(x^x)
and (sqrtx)^(sqrtx).e^(x^2)

and integrate these: cosxln(sinx)/sinx.dx

..and a toughy ((x^1/4)+5)/(x-16).
12. (Original post by Silly Sally)
I am sure integrating arcsinx/getting arcsin x from an integral isn't on the edexcel sylabus for p3
well you didnt have to integrate arcsin for that question, it was simply a matter of algebra.. rearranging x=sin u in terms of u, i.e. u = sin^1 x
At least now you know!
13. (Original post by kimoni)
well you didnt have to integrate arcsin for that question, it was simply a matter of algebra.. rearranging x=sin u in terms of u, i.e. u = sin^1 x
At least now you know!
Thanks - at least it won't annoy me the rest of the day!!!
14. (Original post by Ralfskini)
Some slightly more challenging ones:

ok, differentiate the following: x^(x^x)
and (sqrtx)^(sqrtx).e^(x^2)

and integrate these: cosxln(sinx)/sinx.dx

..and a toughy ((x^1/4)+5)/(x-16).
Is this P3?
15. yeh looks like it, the first 2 chain rule etc arent they?
16. (Original post by Bhaal85)
Is this P3?

It's well within the capabilities of a P3 student (I haven'd done beyond P3)
17. (Original post by Ralfskini)
Some slightly more challenging ones:

ok, differentiate the following: x^(x^x)
and (sqrtx)^(sqrtx).e^(x^2)

and integrate these: cosxln(sinx)/sinx.dx

..and a toughy ((x^1/4)+5)/(x-16).

is the first one

(x^x^x)(x + 2x*ln x) ?
18. (Original post by Ralfskini)
Some slightly more challenging ones:

ok, differentiate the following: x^(x^x)
and (sqrtx)^(sqrtx).e^(x^2)

and integrate these: cosxln(sinx)/sinx.dx

..and a toughy ((x^1/4)+5)/(x-16).
cosxln(sinx)/sinx = cotxln(sinx)

We have d(ln(sinx))/dx = cotx so a substitution of u=ln(sinx) will give

INT cosxln(sinx)/sinx dx = (1/2)[ln(sinx)]^2 + C
19. (Original post by kimoni)
is the first one

(x^x^x)(x + 2x*ln x) ?

No, keep trying.
20. (Original post by mikesgt2)
cosxln(sinx)/sinx = cotxln(sinx)

We have d(ln(sinx))/dx = cotx so a substitution of u=ln(sinx) will give

INT cosxln(sinx)/sinx dx = (1/2)[ln(sinx)]^2 + C

...is the right answer!

your method is probably better but I did it as follows:

let u=sinx
du=cosx.dx

int.cosxlm(sinx)/sinx=int.(lnu/u).du

Integrating by parts gives 1/2(lnu)^2+C.

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