Pls help me on this integral question..square root of (x^2-36).using trig substitution pls.i can do it using hyperbolic substitution of cosh@.i really need your help pls.with including my maths recruit o.you know at times recruits are talented with skills.lol

benbuks: ..my dear, i actually dnt kw what to tell him again,so i quit responding to his post, imagine i wast'nt even referin 2him..bt paul....please lets let him be...lets continue solving., learning frm each oda.nd helpin doz dat r ready 2 learn....

.i love that ben..

Osoayobami: Pls help me on this integral question..square root of (x^2-36).using trig substitution pls.i can do it using hyperbolic substitution of cosh@.i really need your help pls.with including my maths recruit o.you know at times recruits are talented with skills.lol.i love that ben..

...free dat guy abeg...he's own his own..u wan cause anoda problem ni?

Alpha Maximus: alright, here we go:
Let the first term be 'a' and the common difference be 'x'
1st term : a
2nd term : a+x
3rd term : a + 2x
Since the sum of the product of just two terms and 1 give the same answer, we have that:
(a(a+x))+1=(a(a+2x))+1=((a+x)(a+2x))+1
Therefore,
A^2 +ax+1=a^2+2ax+1=a^2+3ax+2x^2+1
Now there are three equations, we equate the first and last, and temporarily isolate the second(because we don't know which is equal to which yet)
a^2+3ax+2x^2+1=a^2+ax+1
Collecting like terms we have
2ax+2x^2=0
2x(a+x)=0
X=0(impossible,since its an arithmetic progression,there has to be a difference) and a+x=0, a=-x(correct)
Since 2ax+2x^2=0, the middle equation is also equal to 0
Thus,
a^2+2ax+1=0
Now substitute 'a=-x' into the middle/second equation
(-x)^2+2(-x)x+1=0
X^2-2x^2+1=0
Factorising, we have:
(X-1)(x-1)=0
X=1(twice)
Now that we have established the common difference of '1', we input it back into the original equations but this time we subtract 1 from all three sides for simplicity and we have:
a(a+1)=a(a+2)=(a+1)(a+2)
We first equate and solve the first and second
a(a+1)=a(a+2)
a^2+a=a^2+2a
Collecting like terms, we have
-a=0
Thus, a=0
Now we simply equate the third equation to 0
(a+1)(a+2)=0
Thus a=-1 and a=-2
We try out a=-2 as the first term of the series a, a+1, a+2 but it does not satisfy the question
But when a=-1 is tested, it satisfies the question(all products of a pair + 1=a perfect square)
Therefore the terms are -1, -1+1, -1+2=-1, 0 , 1
N.B: recall that in my solution , the middle equation was already equal to 0, before I arrived at my final three answers

.use a-d,a,a+d instead my maths recruit who doesn't want to learn from generals.let me be frank with you bro.you have to humble yourself before everybody.that's the only way you can learn.try and read the post of calculusf(x) you will see something called humility and respect there.YOU ARE GREAT BRO.But bridle ur tongue.though you are just typing here but it's obvious that,that's your true character.nobody is an iceland of knowledge.just allow yourself to be corrected

d citizen:
The five star general back online.
Thank God dat i have jst fully recovered from acute fever and fully ready to make contributions.

@d citizen

Thank God that you fully recovered. While you were away, Sire, we your boys and girl couldn't resolve this. So we said make we wait for our only 5 STAR Math General to helep us scatter am.

jackpot: dear Math Generals, pls help me with this little problem.


[size=16pt]By using the Cartesian-paraboloidal coordinate transformation equations:
(x=uv sin φ, y=uv cos φ, z=¹/₂(u²-v²),
show that the 3-dimensional Laplace's equation
F_xx+F_yy+F_zz=0
reduces to
¹/_(u²+v²) (F_uu+F_vv+ ¹/_u F_u+¹/_v F_v)+¹/_(u²v²)F_φφ = 0.[/size]

Expecting your answer, Sire!

@jackpot

Ur question is about partial differential equation and laplace equation. It has been long i lay my hands on engineering mathematics becos i graduated in 2007 but presently working as a production officer in a reputable company. I am jst using my residual knowledge in basic mathematics especially pure math to deal with problems posed by contributors in the house though i have a mechanical engineering background.i may not be able to make contributions on problem relating to higher matrixes e.g upper n lower triangular matrix, guassian jordan elimination matrices, row transformation matrix, fourier series, numerical analysis particularly, runge kuta, power n fiuboren series, simplex tableau, green n stoke theorem, laplace transformation and partial differentiation and ode. Operador d method, jacobian method -all which i consider as engineering math. U can referred ur problems to laplacian, doubledx, benbuck, calculus and jackpot (our one and only female fermat in d house) and others. i admired their ingenuity and sagacity in math. I believe dat they will pave way for the design of underground satellite tru their math ingenuity. If i can lay my hands on borrowed engineering math on schaum outline on pde or a canders or k.a stroud textbook, i will give u answer to ur problem. Ur question u posted was quite simple, i must confess. Single out. Five star math general

Solution to integral square root x2-36

Y=integral sqr x2-36 dx
Let x = 6cosht

dx/dt=6sinht
dx=6sinht dt

Thus, y= integral sqr((6cosht)^2 -36)dx

Y=integral sqr(36(cosh^2t -1))dx

Y=6integral sqr(cosh^2 -1)dx

y=6 integral sinhtdx

But x=6cosht
dx/dt=6sinht
dx=6sinhtdt

Thus,y= integral 36sin^2htdt

Y=36integral sin^2htdt

Y=36 integral (cosh2t-1/2)dt

Y=18 integral (cosh2t -1)dt


Y= 9sinh2t -18t c

Y= xsqr(x2-36)/2 -18cosh-1(x/6) c

. Five star math genral

Osoayobami: Pls help me on this integral question..square root of (x^2-36).using trig substitution pls.i can do it using hyperbolic substitution of cosh@.i really need your help pls.with including my maths recruit o.you know at times recruits are talented with skills.lol.i love that ben..

y=§sqr(x^2-36)dx, let x=6sec@, dx/d@=6sec@tan@, makin substitutn, y=§sqr(36sec^2@-36)*6sec@tan@d@, it simplifies to, y=§sqr(36tan^2@)*6sec@tan@d@ or y=§6tan@*6sec@tan@d@ or
y=§36sec@tan^2@d@ or
y=§36sec@*(sec^2@-1)d@ or
y=§36sec^3@d@-§36sec@d@ we now integrat d first integral by part...
Y=36sec@*tan@-§36tan^2@sec@d@-§36sec@d@
...or
y=36sec@tan@-y-§36sec@d@
2y=36sec@tan@-§36sec@d@...u only have to integrat sec@ and obtain ur answer...

...to integrat sec@=1/cos@, multiply numerato & denominato by cos@ to get cos@/cos^2@ or
cos@/(1-sin^2@), let p=sin@,
dp/d@=cos@, so that §sec@d@=§dp/(1-p^2)=1/2*Log[(1+p)/(1-p)] ...after resolvin 2 partial fractio n integratn...
§sec@d@=1/2*Log[(1+sin@)/(1-sin@)]

Great thread! I can put confusion here with just one simple question on Change of Subject of Formula. Should I bring it?

hod898: Great thread! I can put confusion here with just one simple question on Change of Subject of Formula. Should I bring it?

..be our guest...

Its simple! Here it goes:
S=UT + 1/2(AT^2)

Make t the subject of formula. Hehe.

hod898: Its simple! Here it goes:
S=UT + 1/2(AT^2)

Make t the subject of formula. Hehe.

.haba dis is nt tough na.hmm anyway....do u nid d solution or the final. answer..?

benbuks: .haba dis is nt tough na.hmm anyway....do u nid d solution or the final. answer..?

Bring both please?.

Alpha Maximus:
Here we go,
S=ut+0.5at^2
Bring over 's' to the R.H.S.
Thus,
0.5at^2+ut-s=0
Multiply through by 2 to make sure each co-efficient is a whole number,
At^2+2ut-2s=0
Comparing with the Almighty Formula, we have that:
a=a, b=2u, c=-2s
T=(-b+/- sqrt(b^2-4ac))/2a
T=(-2u+/-sqrt((2u)^2-4(a*-2s))/2a
T=(-2u+/-sqrt(4u^2+8as))/2a
T=(-2u+/-sqrt(4(u^2+2as))/2a
T=(-2u+/-2sqrt(u^2+2as)/2a
T=2(-u+/-1)sqrt(u^2+2as)/2a
The '2's' in the numerator and denominator cancel out each other and we have a final answer of:
T=-u+/-1sqrt(u^2+2as)/a
Oya where are the critics?

hmmm, can u recheck yr working please?

Alpha Maximus:
Here we go,
S=ut+0.5at^2
Bring over 's' to the R.H.S.
Thus,
0.5at^2+ut-s=0
Multiply through by 2 to make sure each co-efficient is a whole number,
At^2+2ut-2s=0
Comparing with the Almighty Formula, we have that:
a=a, b=2u, c=-2s
T=(-b+/- sqrt(b^2-4ac))/2a
T=(-2u+/-sqrt((2u)^2-4(a*-2s))/2a
T=(-2u+/-sqrt(4u^2+8as))/2a
T=(-2u+/-sqrt(4(u^2+2as))/2a
T=(-2u+/-2sqrt(u^2+2as)/2a
T=2(-u+/-1)sqrt(u^2+2as)/2a
The '2's' in the numerator and denominator cancel out each other and we have a final answer of:
T=-u+/-1sqrt(u^2+2as)/a
Oya where are the critics?

..good boy,clap 4urself...i cn see u r showin urself.....if i correct u nw u'll start arguin,so no nid..

hod898: Its simple! Here it goes:
S=UT + 1/2(AT^2)

Make t the subject of formula. Hehe.

na this yeye subject of the formula (quadratic equation in "t" solutioned)question you wan take use put confusion as you insinuated?

SMH.

@jackport..shr d link 2 ur solution,or wot do u think..?

jackpot:
@d citizen

Thank God that you fully recovered. While you were away, Sire, we your boys and girl couldn't resolve this. So we said make we wait for our only 5 STAR Math General to helep us scatter am. Expecting your answer, Sire!

...i hav d solution 2 ur problm...

@Laplacian

am happy. Post the solution, please.

cconcept: any MATLAB guru here?

still waiting for a general to respond

hod898:
hmmm, can u recheck yr working please?

...for what? The right answer is actually the first: -u+1sqrt(u^2+2as)/a.....or is it the format I typed it in?

benbuks: @jackport..shr d link 2 ur solution,or wot do u think..?

acknowledged, Sir. Will do that soon. Let first see what Laplacian's have got!

@Laplacian
let's go there. Hope you won't demand ransom first

post your solution na, pls

cconcept:

still waiting for a general to respond

na advanced program be that. I attended a workshop on the sister program SCILAB, but I must say, I don forget the thing sef. I'd advise you google MATLAB User Groups.

jackpot: na advanced program be that. I attended a workshop on the sister program SCILAB, but I must say, I don forget the thing sef. I'd advise you google MATLAB User Groups.

yes o,i am going gaga because of GPS and galilieo ooo

jackpot: @Laplacian

am happy. Post the solution, please.

first i need to point out 2 u dat in PDE one can take d differental of an eqn b4 proceedin 2 d limt witout fear of error...it is rigorous & not an assumption...wit our former notation; now x=uvsinq, y=uvcosq, z=0.5(u^2-v^2), like b4,
Fx=Fu*Ux+Fv*Vx+Fq*Qx...to get Ux, Vx, and Qx we take d differential of d transformation equations; first we take Log of x=uvsinq,
Logx=logu+logv+logsinq, let d denot delta, so dat
1/x*dx=1/u*du+1/v*dv+1/sinq*cosq*dq, divide thru by dx and proceed 2 d limt;
1/x=1/u*Ux+1/v*Vx+cotq*Qx, similarly for y=uvcosq,
logy=logu+logv+logcosq,
1/y*dy=1/u*du+1/v*dv-1/cosq*sinq*dq, divide thru by dx but this time partial derivativ of y w.r.t.x is zero since x and y are independent variables;
0=1/u*Ux+1/v*Vx-tanq*Qx...again for z=0.5(u^2-v^2),
dz=udu-vdv, dividin by dx;
0=uUx-vVx...from d 3 eqns above solve 4 Ux, Vx and Qx...in d eqns above, divide, respectively, by dy and dz to obtain Uy, Vy, Qy and Uz, Vz, Qz...substitute and differentiat Fx,Fy,Fz respectively...u 'll obtain ur answer as i 've done here after simplification...

Alpha Maximus:
Here we go,
S=ut+0.5at^2
Bring over 's' to the R.H.S.
Thus,
0.5at^2+ut-s=0
Multiply through by 2 to make sure each co-efficient is a whole number,
At^2+2ut-2s=0
Comparing with the Almighty Formula, we have that:
a=a, b=2u, c=-2s
T=(-b+/- sqrt(b^2-4ac))/2a
T=(-2u+/-sqrt((2u)^2-4(a*-2s))/2a
T=(-2u+/-sqrt(4u^2+8as))/2a
T=(-2u+/-sqrt(4(u^2+2as))/2a
T=-2U+/-2SQRT(2U^2+2AS)/2A
T=2(-u+/-1)sqrt(u^2+2as)/2a
The '2's' in the numerator and denominator cancel out each other and we have a final answer of:
T=-u+/-1sqrt(u^2+2as)/a
Oya where are the critics?

..bro,thumbs up....pls once again let us all stop this unecessary enemity here..i learn enough from this thread at least if no one else does..............can someone help me find the 10th term of the expression..(1+x)^15

Cashio: ..bro,thumbs up....pls once again let us all stop this unecessary enemity here..i learn enough from this thread at least if no one else does..............can someone help me find the 10th term of the expression..(1+x)^15

15Combination9