chai....when I was compiling all the questions on this thread. But at least my username has been changed back from ybnl11....stupid nl

mehn it filz gud 2 b bak!

How can we spend all those time preparing for the Mathematics quiz and all the efforts of the Quiz Masters and the participants go down the drain just like that?

let 's try this welcome back question

compute the anti-derivative of [arcsin(lnx) ] dx







A,b,c is a geometric progression (a,b,c are positive integer) such that
a+b+c=26

and a^2+b^2+c^2=364

find the value of a,b,c



asap .

Pls I'd need recommendations on books to buy....on algebra(linear and abstract)....and analysis(real,vector,complex,numerical).....thanks in anticipation...

killsmith: Pls I'd need recommendations on books to buy....on algebra(linear and abstract)....and analysis(real,vector,complex,numerical).....thanks in anticipation...

schum's outline

hmmmmmmmmmmm............if only seun had agree to purchase my new software......this data loss wouldn't av been a case.....*sob*.......re-booking my corner

hmmmm.

It's a pity, almost 50 pages plus our NMQC II thread gone! Let's build it up again gurus.... don't be discouraged!

2nioshine: hmmmmmmmmmmm............if only seun had agree to purchase my new software......this data loss wouldn't av been a case.....*sob*.......re-booking my corner

General 2nioshine, good to be back...you're a software engineer or something? Seun was careless about backing up, that was the problem...I hope he learns from what happened!

Ben the general...I will give your problems a try later in the day! Top of the morning to you sir!

doubleDx: Ben the general...I will give your problems a try later in the day! Top of the morning to you sir!

ok boss ...thanks ..

mathematicians could be crazy sometimes .

1 ..$[arcsin√(2x) /√(1-2x) ] dx

2..$ [ xlnx/(1+x^2)^2 ] dx

Solve d equation:::
3x+y-z=0....*
-2x+y+4z=0...*
4x+y-2z=0....*.

Pls show full working s...

ok

doubleDx: It's pity, almost 50 pages plus our NMQC II thread gone! Let's build it up again gurus.... don't be discouraged!

maybe we shall organize another one.

Yea. Greetings to all my bosses on the thread.
I think I should resume back now that I'm on a forced holiday. Nice questions @benbuks, I'll attempt them tomorrow.

Errmmmm, where are dejt4u, Arithmetic, STENON, Amazing Angel, Alpha Maximus, Smurfy and others? Were they "tsunamised"?

[quote
author=jaryeh] Yea. Greetings to all my bosses on the thread.
I think I should resume back now that I'm on a forced holiday. Nice
questions @benbuks, I'll attempt them tomorrow.

Errmmmm, where are dejt4u, Arithmetic, STENON, Amazing Angel, Alpha
Maximus, Smurfy and others? Were they "tsunamised"? [/quote] No ooo....Here is ur Girl o....Goodevening n I hope you are doing great?......

STENON: No ooo....Here is ur Girl o....Goodevening n I hope you are doing great?......

Sure. How about you?

jaryeh:
Sure. How about you?

Yes, Same.......Hope u'r enjoyin ur forced brk?

efficiencie: mehn it filz gud 2 b bak!

try my question$ boss.

tnx

jaryeh: Yea. Greetings to all my bosses on the thread.
I think I should resume back now that I'm on a forced holiday. Nice questions @benbuks, I'll attempt them tomorrow.

Errmmmm, where are dejt4u, Arithmetic, STENON, Amazing Angel, Alpha Maximus, Smurfy and others? Were they "tsunamised"?

Tsunamised? I thinketh not.

STENON: Yes, Same.......Hope u'r enjoyin ur forced brk?

Forced break! Mehn, it has not been funny at all. House sweet sha, na chores spoil am. You're at school right?

Alpha Maximus:
Tsunamised? I thinketh not.

Boss, long time. How and where have you been?

jaryeh: Yea. Greetings to all my bosses on the thread.
I think I should resume back now that I'm on a forced holiday. Nice questions @benbuks, I'll attempt them tomorrow.

Errmmmm, where are dejt4u, Arithmetic, STENON, Amazing Angel, Alpha Maximus, Smurfy and others? Were they "tsunamised"?

present sir..what it do!

benbuks: let 's try this welcome back question

compute the anti-derivative of [arcsin(lnx) ] dx



asap .

Integral [arcsin(lnx)]dx=? Now, applying the method of integration by part. Let arcsin(lnx)=u and dx=dv. Integrating both dv and dx will yield v=x. To differentiate arcsin(lnx) let lnx=w, therefore arcsin(lnx)=arcsinw.
Since arcsin(lnx)=arcsinw=u; >> du/dw =1/[sqr(1-w^2)].
From w=lnx, >> dw/dx =1/x.
Therefore du/dx =[1/[sqr(1-w^2)]] * 1/x.
Since w=lnx, >>> du/dx = [1/1-sqr(lnx^2)] * 1/x.
We can now say that du=dx/[xsqr(1-lnx^2].
Now integral udv=uv- integral vdu
>> integral arcsin(lnx) = xarcsin(lnx) - integral xdx/xsqr(1-lnx)^2)+c.
To integrate dx/sqr(1-lnx^2) let lnx=sinu
Therefore dx/x = cosudu >> dx=xcosudu.
Now integral 1/sqr(1-lnx^2)=integral [1/sqr(1-sin^2 u)]*xcosudu.
Which equal to integral xcosudu/sqr(1-sin^2 u).
From sin^2 u + cos^2 u = 1 >>> 1-sin^2 u=cos^2 u.
We now have integral xcosu/sqrcos^2 u = xcosu/cosudu.
Therefore integral dx/sqr(1-lnx^2) = integral xdu.
Since lnx=sinu >> x=e^sinu.
Integral dx/sqr(1-lnx^2)= integral e^sinudu
From maclaurin series e^x=1+x+x^2/2+.......
>>e^sinu=1+u+u^2u/2!+......
Integral e^sinudu=u+u^2/2+u^3/6+...
Now lnx=sinu
>>u=arcsin(lnx)
Integral e^sinudu= arcsin(lnx)+1/2[arcsinlnx]^2+1/6[arcsinlnx]^3+......
Combining the results
Integral arcsin(lnx)dx= xarcsin(lnx)+arcsinlnx+1/2[arcsinlnx]^2+1/6[arcsinlnx]^3

I hope I'm not wrong

Sir Chides: Solve d equation:::
3x+y-z=0....*
-2x+y+4z=0...*
4x+y-2z=0....*.

Pls show full working s...

3x+y-z=0................(i)
-2x+y+4z=0............(ii)
4x+y-2z=0................(iii)
(i)-(ii)
5x-5z=0
5x=5z
X=z......................(iv)
Sub eqn (iv) into (iii)
4z+y-2z=0
2z+y=0
Y=-2z
Choosing an arbitrary solution for z
Let z=1
>>>x=1 and y=-2(1)=-2
Therefore x=1,y=-2,z=1.

benbuks: 1 ..$[arcsin√(2x) /√(1-2x) ] dx

I want to believe $ means integral.
Integral arcsin(sqr2x)/sqr(1-2x)
Using the method of integration by parts.
Let u=arcsin(sqr2x) and dv=1/sqr(1-2x)dx
Now du=dx/sqr(1-(sqr2x)^2) and v=arcsin(sqr2x)
It now implies that du=dx/sqr(1-2x)
Integral udv =uv- integral vdu
>> integral [arcsin(sqr2x)/sqr(1-2x)]=
arcsin(sqr2x)*arcsin(sqr2x) - integral [arcsin(sqr2x)dx/sqr(1-2x)]
Collecting like terms
2[Integral arcsin(sqr2x)/sqr(1-2x)]= sin^-2(sqr2x)
Therefore integral [arcsin(sqr2x)/sqr(1-2x)]=
[Sin^-2sqr(2x)]/2 + c

See works ere wey i no fit solve... Chaii,i don suffer ooo.... Ehen abeg make una elp me solve dis palasa quest



Integal Sec(X)dx

oladistinct:
I want to believe $ means integral.
Integral arcsin(sqr2x)/sqr(1-2x)
Using the method of integration by parts.
Let u=arcsin(sqr2x) and dv=1/sqr(1-2x)dx
Now du=dx/sqr(1-(sqr2x)^2) and v=arcsin(sqr2x)
It now implies that du=dx/sqr(1-2x)
Integral udv =uv- integral vdu
>> integral [arcsin(sqr2x)/sqr(1-2x)]=
arcsin(sqr2x)*arcsin(sqr2x) - integral [arcsin(sqr2x)dx/sqr(1-2x)]
Collecting like terms
2[Integral arcsin(sqr2x)/sqr(1-2x)]= sin^-2(sqr2x)
Therefore integral [arcsin(sqr2x)/sqr(1-2x)]=
[Sin^-2(2x)]/2 + c

bozz i tried solving dis qwest by follown ur steps bt i dint gt it.... If dv= 1/sqr(1-2x)?? Den V can nv be arcsin sqr(2x)...

layez:




bozz i tried solving dis qwest by follown ur steps bt i dint gt it.... If dv= 1/sqr(1-2x)?? Den V can nv be arcsin sqr(2x)...

If dv=1/sqr(1-2x)dx
Integrating both sides
Integral dv= integral 1/sqr(1-(sqr2x)^2)dx
This is identical to the standard integral 1/sqr(1-x^2)dx which will give arcsinx
Therefore integral 1/sqr(1-(sqr2x)^2) will yield arcsin(sqr2x).
Hope that helps