₦airaland Forum

efficiencie: i guess ur encouragement paid off, i'v been trying sir chides' questn bt d soln az been eluding me...bt dis tym, i gues i nailed it


∫(1/(cosk + cosx))dx

Let cosk=1/a

= a∫(1/(1 + acosx))dx

= a∫(1/(cos^2(x/2) + sin^2(x/2) + acos^2(x/2) - asin^2(x/2) ))dx

= a∫(1/(cos^2(x/2) + acos^2(x/2) + sin^2(x/2) - asin^2(x/2)))dx

= a∫(1/((1+a)cos^2(x/2) + (1-a)sin^2(x/2)))dx

= a∫sec^2(x/2)dx/((1+a) + (1-a)tan^2(x/2)))

= (a/(1+a))∫sec^2(x/2)dx/(1 + {√[(1-a)/(1+a)].tan(x/2)}^2)

Let tan(u/2)=√[(1-a)/(1+a)].tan(x/2)

(1/2)sec^2(u/2)du=(1/2)√[(1-a)/(1+a)].sec^2(x/2)dx

√[(1+a)/(1-a)].sec^2(u/2)du= sec^2(x/2)dx

hence
(a/(1+a)) ∫sec^2(x/2)dx/(1 + {√[(1-a)/(1+a)].tan(x/2)}^2)

Becomes
= (a/(1+a)) ∫√[(1+a)/(1-a)].sec^2(u/2)du /(1 + tan^2(u/2))

= (a/(1+a)). √[(1+a)/(1-a)]∫du

= (a/(1+a)) .√[(1+a)/(1-a)]u

= {2a/√[(1+a).(1-a)]}. arctan{√[(1-a)/(1+a)]tan(x/2)}
Where 1/a=cosk

over to the auditors

Like u said, let cosk=1/a
Let $=integral sign
>>>$[1/(cosk+cosx)]dx=a$[1/(1+acosx)]dx
Which is the same as a$[1+acosx]^(-1) dx
Using binomial expansion
Let C=combination
a$[1+acosx]^(-1) dx=a$[1+ (-1)C1acosx+(-1)C2acos^2 x+....]dx
Now -1C1=-1 and -1C2=1
>> a$[1+cosx]^(-1) dx=a$[1-acosx+a^2cos^2 x +....]dx
=a[x-asinx+$(a^2 cos^2 x)dx+...] +c
Now cos^2 x=[1+cos2x]/2
>>a$[1+cosx]^(-1) dx=a[x-asinx+(a^2 x)/2 + (a^2 sin2x)/4]+.... +c
Since cosk=1/a >>> a=1/cosk.
Therefore $[1/(cosk+cosx)]=a[x - asinx + (a^2 x)/2 + (a^2 sin2x)/4 + ..... ] + c.

layez: i must commend ur effort by typing d soln but u almost made my head go haywire cos of d way u solved... Aiit,i grab cos i did it too.. Firstly, let * = Exp
then, Tn=ar*n-1 so you made a mistake dere....

Thumbs up

Thanks bro, it was a typo error

benbuks: 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 .

Let us consider the expansion of [a+b+c]^2
[a+b+c]^2=a^2+b^2+c^2+2[ab+ac+bc]
Now a+b+c=26 and a^2+b^2+c^2=364
>>26^2=364+2[a+b+c]
676-364=2[ab+ac+bc]
ab+ac+bc=312/2=156
we now have ab+ac+bc=156.............(i)
and a+b+c=26....................(ii)
Since the progression is geometric
Tn=ar^[n-1]
>>b=ar and c=ar^2 since b and c are 2nd and 3rd term respectively
Substitute these into equation (i) and (ii)
a+ar+ar^2=26 >>>a[1+r+r^2]=26.......(iii)
also a^2r+a^2r^2+a^2r^3=156
>>>a^2r[1+r+r^2]=156.........(iv)
Divide iv by iii
[a^2r(1+r+r^2)]/[a(1+r+r^2)]=156/26
>>>ar=6 therefore r=6/a
Now b=ar=6
c=ar^2=[ar]r=6r
From a+b+c=26
a+6+6r=26
a+6r=20
Now r=6/a
a+6[6/a]=20
a+36/a=20
>>> a^2+36=20a
a^2-20a+36=0
Solving the above quadratic equstn
a=18 and 2
using a=2
Since c=ar^2
Now r=6/a=6/2=3
>>>c=2*3^2=2*9=18
Therefore a=2, b=6 and c=18.

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

Integral xlnx/[1+x^2]^2dx
Integration by part
U=lnx >> du=dx/x
dv=x/[1+x^2]^2dx
Let p=1+x^2
dp=2xdx >>dx=dp/2x
Therefore v=integral [x/p^2]*dp/2x
=integral 1/2[p^-2]dp
=-1/2p
Since p=1+x^2
>>>v=-1/2[1+x^2]
Now integral udv=uv- integral vdu
>>integral xlnx/[1+x^2]^2dx= -lnx/2[1+x^2]- 1/2integral[-1/1+x^2]*1/xdx
=-lnx/2(1+x^2)+1/2integral[1/x(1+x^2)]dx
Resolving 1/x(1+x^2) into partial fraction
1/x(1+x^2)=1/x -x/1+x^2
We now have -lnx/2(1+x^2)+1/2integral[1/x -x/1+x^2]dx
We can adjust x/1+x^2 to 1/2[2x/x^2+1]
>>integral xlnx/[1+x^2]^2=-lnx/2(1+x^2)+1/2lnx-1/4ln[x^2+1].

Ogunbiyi olalekan
Oau ile-ife
Mechanical Engineering
500 level
Maths, physics
oladistinct@yahoo.com

benbuks: nice work @ oladistinct .

Yes boss

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

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

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: 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

maafen: ODE FOOL It was an intentional hit. Go ãпԃ sleep if yόϋ ve got nothing †☹ say

Stop ranting, you are free to kill yourself

Rossikk: YOU'RE VERY NAIVE. OR BLIND.

Same here

Matuidi is not a dirty player. He went for the ball but his tackle was late, it can happen to anyone.

chronique: To everyone who says there is no God: all I need y'all to do is to prove to me how man was created/came into existence. If you can tell me how man's flesh was made,how your respiratory system was designed,how your brain was programmed,how the concept of blood being the soul of a man,was achieved,etc;I'd believe there's no God. It's only a fool that will claim He knows for sure,and can give you adequate info on what happened in a place He has never been to. It's only a fool that will come out and contradict historical events that occurred before his/her birth. It's only a fool that will wake up and say a way of life that has been observed by thousands of generations in the past,is false,without giving concrete evidences to prove it. You cannot explain how you came into existence,yet you claim there's no creator or God? Smh. The difference between the fool and the wise is that,the wise will believe there's God,only to find out at the end that there is no God. But the foolish will assume there's no GOD,only to discover at the end,that there is a God that created heaven and earth.

I understand you quite well; because the human body is complex and you can't explain how it was formed gave rise to your god hypothesis. It's almost impossible to know what transpired many years ago when life first grazed our planet but i tell you, the big bang theory is one of those theories that try to explain how the universe and life came into being. It is not a perfect theory to me though but it makes sense more than the theory of creation in the bible where god put the carth before the horse by creating earth before the sun and go on to create the first man with dust and breath of life: your god must be a magician.

peterphd: If atheism is a religion then abstinence is a se.xual position.

#GBAM

bigblxd: @ OP, with all sense of humility,threads like this gives me nightmares! Why on earth will there not be a God! Pls God is using your mom to give you one more chance; pls & pls open up your heart. Faith is now; salvation is today; Jesus loves you!

Op, with all sense of humility, Allah is using your mum to give you one more chance; pls & pls open your heart.Faith is now; salvation is today; Mohammed loves you.

showby: What do you expect from an atheist, a mumu.

What do you expect from a theist, a deluded mind.

ahmstrng: ATHEISM IS A RELIGION OF ITS OWN

Far from the truth, stop seeing what others hold on to as a replica of what you are engaged in.

ifeness: How did you come to know this?

He is the P.A. of his god

Teeted: Sorry!
Ndo!
Pele!
Doo!
Eeyaaa!
Mmmmmn!!!!

same here brother

holamiday: So, what are you saying in essence? Are you part of the atheists?

you want to put me on gej's list?

[quote author=Okija_juju]Gbam...


I had no idea where I came from.. If I could have planned for this life, I'm sure Nigeria would not have been my first choice, nor Africa and my Parents would most likely be Bill Gates or someone wealthy like that.. [/quote]

holamiday: No wahala. Sebi we'll all die one day. You people think we christians don't know what we are talking about when we talk of this Jesus. He is the image of the invisible God.

Thank your god for you, you said it yourself, invisible god but I'll help you a little, it should have been imaginary god. Trust me my brother, you christians don't know what you are talking about. when you die you can't know where you will be, because you don't know where you were before you were born.

lizbum: Ask God to open d eyes of ur undrstnding to kw him in person

Why on earth should u ask the ever knowing and ever present god for that? god should be smarter than that now, after all we didn't ask him to create us in the first instance yet he created us, why should we now open our heart and look forward to him revealing himself to us. If indeed he is the almighty, he should have come up with a plan to clear the air all this while. Maybe he is now in extinction sha

holamiday: I'm talking about the great grand mother of all failures: eternal condemnation.

Now I get your drift, eternal condemnation? how? where? You mean eternal condemnation to people that do not believe in that god of yours, I laugh you. Think about this; what if Mohammed, Budha or Krishna happen to be the right way to the almighty; alas, we will enjoy that eternal condemnation together, if it's real anyways.

holamiday: These words may one day stand against you. Without Jesus, you will fail.

Are u saying people following Jesus don't experience failure? Failure is one of the cycles of life, it has nothing to do with the son of your imaginary sky daddy.

adezjamz: Guy U̶̲̥̅̊ never die ?

Padi I still dey stand gidigba for hia o.

[quote author=Okija_juju]Also from the mouth of Jesus

Matthew 23:17 "You blind fools! Which is greater: the gold, or the temple that makes the gold sacred?"


And...


Luke 11:40 You foolish people! Did not the one who made the outside make the inside also?




Judge Jesus by his words...


_{Dont ask me, Ask Jesus.. The gods are wise..}
[/quote]Jesus should and must be in hell: use the medallion to kill the medallion.

Guys ar just too crazy out here, imagine mazaje forming Mr muslim; lwkmd

Ajibam: Federal University of Agriculture, Abeokuta

Nah, obafemi awolowo university, ile - ife

I'm gonna miss LB

rationalmind: In addition, what exactly is the rationale in banning someone for using an "offensive" monicker when he'll be back with that same monicker when the ban expires.

Wouldn't it be more sensible if he's told via a pm to deactivate the account and register with a new username?

You took that out of my mouth. What seun wants to achieve with the ban is what i don't seem to understand since he'll be back as the same Mr. Troll after the ban.