layez: 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

can do this in multiple ways , lets use a simple one

now, considering the function , we re-write as

secx .(tanx +secx)/(tanx +sec x )

=( secx.tanx+sec^2 x) /(tanx +secx )

now taking integral

= $ [(sec^2 x +tanx.secx) /(tanx +sec x) ]dx

by inspection , we see that the denominator (tanx +secx) is the exact derivative of the numerator

, since D(tanx + secx) = sec^2 x + tanx.secx

=> $ du/u = ln |u| + c ..where u = tanx + secx

thus we have $secxdx = ln|tanx+secx|+ c

hope you get ..?

nice work @ oladistinct .

oladistinct:
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

yaga,i gt

dejt4u:
present sir..what it do!

oga mi, I gentle o. How have you been?

benbuks: nice work @ oladistinct .

Yes boss

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].

I just want to say thank you to all the great mathematicians here, especially @ Benbuks . Wrote my Mth 102, exam today and it was great. Merci Beaucoup

jaryeh:
oga mi, I gentle o. How have you been?

m very good..tnx..hope u ar enjoyin dis your 'Omole cum madam durosimi' compulsory break

Olarewajub: I just want to say thank you to all the great mathematicians here, especially @ Benbuks . Wrote my Mth 102, exam today and it was great. Merci Beaucoup

wow..that's nice bro.

we give all glory to God almighty for wisdom , knowledge & understanding , like one of my bosses will say " A CANDLE LOSSES NOTHING BY LIGHTENING ANOTHER " try also/always to impact / give a helping hand the little you know/have to others in need , don't be selfish !! the more you teach , the more you learn & the more your reward/blessing

kudos to you all my mathematical friends/generals , your efforts here is not / shall never be in vain .


wish you the best in your exams bruv. trust it shall be glorious .

1luv.

dejt4u:
m very good..tnx..hope u ar enjoyin dis your 'Omole cum madam durosimi' compulsory break

Ha! I dey manage am o. I have no option.

jaryeh:

Ha! I dey manage am o. I have no option.

abi na.. That is OAU for u ooo my guy..na so 5 yrs go turn to 6 b dat ooo without havin extra...welcome on board

dejt4u: abi na.. That is OAU for u ooo my guy..na so 5 yrs go turn to 6 b dat ooo without havin extra...welcome on board

Sure. We don gba kamu o.
I hope you're enjoying life after school sir.

Four persons are chosen at random from a group containing 3 men , 2women & 3children show that the probability that exactly two of them are children is 10/21






A five digit number is formed by using 0,1,2,3,4&5 without repetition . find the probability that the number is divisible by 6 ..

asap..

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.

oladistinct:
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.

great work sir..

oladistinct:
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
>>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.

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

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: let 's try this welcome back question

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

Given
I=∫(arcsin(lnx))dx
Let u=arcsinlnx
sinu=lnx
cosu.du=dx/x
dx=e^sinu.cosu du
I=∫(u. e^sinu.cosu) du
I=u. e^sinu – ∫(e^sinu) du
For values of sinu defined as: -1<sinu<1, I can apply the taylor’s rule such that:
I=u. e^sinu – ∫(1+sinu+(1/2!)sin^2(u)+(1/3!)sin^3(u)+(1/4!)sin^4(u)+…) du
Since for higher powers n of sin^n(u) the integral peters out I stop, albeit arbitrarily at n=4
Hence
I≈u. e^sinu – ∫(1+sinu+(1/2!)sin^2(u)+(1/3!)sin^3(u)+(1/4!)sin^4(u)) du
I≈u. e^sinu – ∫(1+sinu+(1/2!)( (1/2) – (1/2)cos2u)+(1/3!)( (3/4)sinu – (1/4)sin3u)+(1/4!)( (3/8 ) – (1/2)cos2u + (1/8 )cos4u)) du
I≈u. e^sinu – ∫(1+sinu+( (1/4) – (1/4)cos2u)+( (1/12)sinu – (1/24)sin3u)+( (1/64) – (1/48 )cos2u + (1/192)cos4u)) du
I≈u. e^sinu – [u – cosu + (u/4) – (1/8 )sin2u – (1/12)cosu + (1/72)cos3u + (u/64) – (1/96)sin2u + (1/768 )sin4u]
I≈u. e^sinu – u + cosu – (u/4) + (1/8 )sin2u + (1/12)cosu – (1/72)cos3u – (u/64) + (1/96)sin2u – (1/768 )sin4u
I ≈ -81/64 + u. e^sinu + (13/12)cosu + (13/96)sin2u – (1/72)cos3u – (1/768 )sin4u
Put the value of u
I ≈ -81/64 + x.arcsinlnx + (13/12)√(1-(lnx)^2) + (13/48)lnx√(1-(lnx)^2) – (1/36)(1-(lnx)^2)) √(1-(lnx)^2) – (1/72)√(1-(lnx)^2) – (1/768 )sin[4(arcsinlnx)]
I left the last term unexpanded…this question sef!
Over to the auditors

benbuks: Four persons are chosen at random from a group containing 3 men , 2women & 3children show that the probability that exactly two of them is 10/21






A five digit number is formed by using 0,1,2,3,4&5 without repetition . find the probability that the number is divisible by 6 ..

asap..

Four persons are chosen at random from a group containing 3 men , 2women & 3children show that the probability that exactly two of them ARE WHAT is 10/21 incomplete question sir

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

Pls show full working s...

see pic below sir:

efficiencie:

Four persons are chosen at random from a group containing 3 men , 2women & 3children show that the probability that exactly two of them ARE WHAT is 10/21 incomplete question sir

"children"

modified ..

*Goodluck Jonathan-President of Nigeria
(Zoologist)
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(Civil Engineer)
*Pastor E.A. Adeboye-General Overseer, RCCG
(Mathematician )
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(Mathematician). *Nkem owoh popularly known
as Osofia Nigerian actor and comedian-(Engin
eer).
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WINNERS CHAPEL (Architect)
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(Microbiologist)
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Engineer) & Many more.
FRIENDS, U need GRACE not human
connection. U need GOD'S Favor not much
friends. U need GOD'S back up not human
standard.
In Proverbs, 23 vs 4. He says. Labour not to be
rich and cease from your own wisdom.
Now, GOD visited Abraham, the farmer and
turns him to generational blessings.
He visited Moses, the stammerer and turns him
generational giant.
BEFORE THIS YEAR IS OVER, IF ONLY U HAVE
FAITH & BELIEVE, U shall become a living
wonders to yourself, your household, your state
& your Nation in Jesus mighty name!
Ur destiny shall not turn to rags in Jesus name.

Dear Math Generals, Lieutenants and Sergeants

Evaluate the limit: lim _{x -> 0} (sin x)^{sin x}

5Star tags: doubleDx, Laplacian, Richiez, benbuks, efficiencie, dejt4u, Amazing Angel, STENON, Mathematical Scientists and Engineers.

[quote
author=jackpot]Dear Math Generals, Lieutenants and Sergeants

Evaluate the limit: lim _{x -> 0} (sin x)^sin
x

5Star tags: doubleDx, Laplacian, Richiez, benbuks, efficiencie, dejt4u,
Amazing Angel, STENON, Mathematical Scientists and Engineers. [/quote] My Boss!!...I salute you o.....Awwwww.....!!!.............Brb ooo....


Goodpm to you.......

jackpot: Dear Math Generals, Lieutenants and Sergeants

Evaluate the limit: lim _{x -> 0} (sin x)^{sin x}

5Star tags: doubleDx, Laplacian, Richiez, benbuks, efficiencie, dejt4u, Amazing Angel, STENON, Mathematical Scientists and Engineers.

HMM dis my madam don show again oo

solution below , my boss don devour d tin

i"ll try alternative solution sha..


most limits of indeterminate forms. yields 1 .

jackpot: Dear Math Generals, Lieutenants and Sergeants

Evaluate the limit: lim _{x -> 0} (sin x)^{sin x}

5Star tags: doubleDx, Laplacian, Richiez, benbuks, efficiencie, dejt4u, Amazing Angel, STENON, Mathematical Scientists and Engineers.

Lim (sinx)^(sinx) as x→0
Let y=(sinx)^(sinx)
lny=sinx.lnsinx
lny=lnsinx/cosecx
Which results in -∞/∞ as x→0
Limlny=lim(lnsinx/cosecx)
Limlny=lim((dlnsinx/dx)/(dcosecx/dx))
Limlny=lim(cotx/-cosecxcotx)
Limlny=-lim(1/cosecx)
Limlny=-lim(sinx)
lny=0 as x→0
y=1
Hence
Lim (sinx)^(sinx)=1 as x→0

efficiencie:


Lim (sinx)^(sinx) as x→0
Let y=(sinx)^(sinx)
lny=sinx.lnsinx
lny=lnsinx/cosecx
Which results in -∞/∞ as x→0
Limlny=lim(lnsinx/cosecx)
Limlny=lim((dlnsinx/dx)/(dcosecx/dx))
Limlny=lim(cotx/-cosecxcotx)
Limlny=-lim(1/cosecx)
Limlny=-lim(sinx)
lny=0 as x→0
y=1
Hence
Lim (sinx)^(sinx)=1 as x→0

just on point bruv. your great man..

dp

jackpot: Dear Math Generals, Lieutenants and Sergeants

Evaluate the limit: lim _{x -> 0} (sin x)^{sin x}

5Star tags: doubleDx, Laplacian, Richiez, benbuks, efficiencie, dejt4u, Amazing Angel, STENON, Mathematical Scientists and Engineers.

Yes, it's = 1

I think general efficiencie solved it correctly!...I will check out later. 1luv

I'm a little busy now!

Someone posted the below question, I quess Sir Chides and I think the doctors in the clinic need to diagnose and profer 'curative' measures for the problem:

Evaluate: ∫dx/(cosk + cosx)

how much glucose syrup with 20 % concentration has to be mixed with 100kg glucose syrup with 40 % concentration so that the mixture will have 36 % glucose?

efficiencie:


Lim (sinx)^(sinx) as x→0
Let y=(sinx)^(sinx)
lny=sinx.lnsinx
lny=lnsinx/cosecx
Which results in -∞/∞ as x→0
Limlny=lim(lnsinx/cosecx)
Limlny=lim((dlnsinx/dx)/(dcosecx/dx))
Limlny=lim(cotx/-cosecxcotx)
Limlny=-lim(1/cosecx)
Limlny=-lim(sinx)
lny=0 as x→0
y=1
Hence
Lim (sinx)^(sinx)=1 as x→0

I am having problems with the bolded. How can you equate "limit" both sides and cancel out?

Next, that LimIny is as y tends to what?