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Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 11:37pm On Feb 08, 2015

Evaluate ${(1/lnt)}dt.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 1:30am On Jan 14, 2015

Arithmetic:
1). $[1 / {3^x + e^-x}]dx.
2). $[log(1-x)/e^-2x]dx. Thanks...

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 1:11am On Jan 12, 2015

1). $[1 / {3^x + e^-x}]dx.
2). $[log(1-x)/e^-2x]dx. Thanks...

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 1:02am On Jan 12, 2015

Laplacian:
recall, sec²y=1+tan²y.
From the LHS,
sec²x-1=1+2(sec²y-1) or
sec²x=2sec²y or
cos²y=2cos²x or
cos²y=(2cos²x-1)+1, does the expression in the bracket look familiar? *****

That's cos2x.
So, cos²y=cos2x+1
or
cos²y-1=cos2x
can u finish it from here? Hop it helps!

Sure sir!.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 2:57am On Jan 10, 2015

If tan²x = 1 + 2tan²y. Show cos2x + sin²y = 0. Thanks…

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 4:28am On Aug 31, 2014

benbuks: Integrate arctan([(1-x)/(1+x)]dx

asap.

§tan^-1{(1-x)/(1+x)}dx.
:::SOLUTION:::
Integrating by part, we have;
u=tan^-1{(1-x)/(1+x)}, v=x, dx=-du(1+x²),
§udv=uv-§vdu,
=tan^-1{(1-x)/(1+x)}.x-§-x/(1+x²)dx.
=xtan^-1{(1-x)/(1+x)}+§x/(1+x²)dx.
For §x/(1+x²)dx, we say, let u=x², dx=du/2x. Substituing, we have;
(1/2)§1/(u+1)du.
1/2ln(u+1)+c, recall, u=x².
=1/2ln(x²+1)+c.
.^.. => xtan^-1{(1-x)/(1+x)} + 1/2ln(x²+1) + c.

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Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 3:17pm On Aug 23, 2014

@ kendzyma , u can only solve using trig. subs.
§¥(x^2-4)^3dx, ¥ rep. sq.root.
SOLUTION:
Let x=2sec@, dx=2tan@sec@d@.
§¥(4sec^2@-4)^3dx.
§{¥4(sec^2@-1)}^3dx.
Recall, 1+tan^2@=sec^2@.
§(2tan@)³.2tan@sec@d@.
16§tan⁴@sec@d@.
tan⁴@=(sec²@-1)².
16§(sec⁴@+1-2sec²@)sec@d@.
16[§(sec⁵@d@+§sec@d@-2§sec³@d@].
Taking one by one, we have;
§sec³@d@. Integrating by part,
u=sec@,v=tan@,du=tan@sec@d@.
=tan@sec@-§tan²@sec@d@.
§(sec²@-1)sec@d@=§sec³@d@-§sec@d@.
Let I=§sec³@d@.
I=tan@sec@-I+§sec@d@,
§sec³@d@= {tan@sec@+ln|sec@+tan@|}/2+c.
§sec@d@=ln|sec@+tan@|.
§sec⁵@d@=§sec³@.sec²@d@. Integrating by part,
§sec⁵@d@=sec³@tan@/4+3tan@sec@/8+3ln|sec@+tan@|.
Substituting, into the real integral and also x=2sec@.

x³¥(x²-4)/4-5x¥(x²-4)/2+6ln|{x+¥(x²-4)}/2|.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 1:14pm On Aug 22, 2014

benbuks: --------SOLUTION--------------
(a)

= 3(i¹⁰⁴ )i + 5(i⁷² )i -8(i² )²⁵ -2(i¹¹⁸ ) i

=>3i(1) +5i(1) - 8(-1) -2i(-1)

= 10i +8

(b)

by difference of two squares.
we have
[(3i-5)² ]² - [(2-i)² ]²

=>(9i² -30i+25)² -(4-4i+i²)²

=>(16-30i)² - (3-4i)²

=> (256-960i +900i² )-(9-24i+16i² )
=>-544-960i +7 +24i

hence we have ,

-537 -936i

@boss, (b), can be solved also by DE MOIVRE'S Theorem... My opinion.

Education › Re: 2013/2014 Obafemi Awolowo University ::ASPIRANTS:: by Arithmetic(m): 8:49am On Aug 18, 2014

@ boss Kendzyma , I do wish to ask, what's your department?.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 3:38pm On Aug 15, 2014

benbuks: x=9/4

prove by solution if u can bro...

i think newton-raphson cud b d guy 2 call ..

what thinketh thou .?

Oh!, I'm very sorry sir, I made a silly mistake along the way.
:::SOLUTION:::
18¥x=729^¥x-1.
Let a=¥x, we have; NB:¥ rep. square root.
18a=729^a-1,
=>3².2a=3^6(a-1).
=>2a=3^{6(a-1)-2}.
=>2a=3^(6a-6-2).
=>3^{6a-8}-2a=0.
Using Binomial Expansion, we have;
(1+2)^6a-8-2a=0.
=>1+2{6a-8}+2².{6a-8}(6a-9)/2+...-2a=0.
=>1+12a-16+2(36a²-102a+72)-2a=0.
=>1+12a-16+72a²-204a+144-2a=0.
On simplifying, we have;
=>72a²-194a+129=0.
Solving the affected quadratic eqn.
=>a=1.5 or 1.194444.
=>a=3/2 or 43/36.
We take the value of a=3/2, which satisfies the equation.
Recall a=¥x,
=>x=a².
.^..x=(3/2)²=9/4.
Newton Raphon's iterative method will also do.#stillyourboy

Education › Re: Calculus Mental Challenge (CMC) WINNER ::::::::::::DRNIYI4U:::::::::: by Arithmetic(m): 11:41am On Aug 15, 2014

Hmm, interesting. Congrats to the winner.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 12:22pm On Aug 13, 2014

1Book: Hmmmm...Ur solution is quite wrong! ...
use linear differential equation method.
what was ur integrating Factor?

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 12:22pm On Aug 13, 2014

1Book: Hmmmm...Ur solution is quite wrong! ...
use linear differential equation method.
what was ur integrating Factor?

Then explain boss.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 12:19pm On Aug 13, 2014

benbuks: 18 √x=729^(√x-1)
solve..

x is a very small value which ranges from 0 to infinity. Hence x approaches infinity.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 1:29am On Aug 13, 2014

Question:
Solve:dy/dx+2ytanx=sinx. By 1Book , from Calculus Mental Challenge(CMC).

:::SOLUTION:::.
Integrating both sides,
$(dy/dx+2ytanx)dx=$sinxdx.
$dy/dx.dx+2y$tanxdx=$sinxdx.
$dy+2y(-lncosx)=-cosx+c
NB: y is taken as a constant in $2ytanxdx.
y-2ylncosx=-cosx+c.
y(1-2lncosx)=-cosx+c.
.'.y=(-cosx+c)/(1-2lncosx). Which is the required solution.

Education › Re: Calculus Mental Challenge (CMC) WINNER ::::::::::::DRNIYI4U:::::::::: by Arithmetic(m): 8:37pm On Aug 12, 2014

Soluton:::
$dx/(x^2-1)
Resolvng 1/(x^2-1) into partial fraction, we have;
1/2[{1/(x-1)}-{1/(x+1)}]. 1/2${1/(x-1)}-{1/(x+1)}dx.
1/2[$1/(x-1)dx-$1/(x+1)]dx.
1/2[ln(x-1)-ln(x+1)]+c.
=>1/2[ln{(x-1)/(x+1)}]+c.

Education › Re: Calculus Mental Challenge (CMC) WINNER ::::::::::::DRNIYI4U:::::::::: by Arithmetic(m): 10:59pm On Aug 11, 2014

e^arctanx/(1+x^2). You may nt decide to add it.

Education › Re: Calculus Mental Challenge (CMC) WINNER ::::::::::::DRNIYI4U:::::::::: by Arithmetic(m): 10:54pm On Aug 11, 2014

Oh!, tight schedule and no more entrance exam?. cry

Education › Re: Calculus Mental Challenge (CMC) WINNER ::::::::::::DRNIYI4U:::::::::: by Arithmetic(m): 3:48pm On Aug 11, 2014

Have you seen my mail?.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 5:52pm On Aug 10, 2014

benbuks: i can sight my great maths lords in the house

@ kendyzma ,arithmetic PatEinstEin

i bow 4 una ooo

naso una like maths like madt .? hmm

i

You are d boss, no flattering sir.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 3:21pm On Aug 10, 2014

benbuks: 1)

There are two answers ->a natural number(ofcourse x=1€N),
->x=[{(¥a+¥b)²+4}/{(¥a-¥b)²-4}]. ¥ still denotes sq.root. Thanks.
No need for no 2'solution. I just gave d ques. except if u have an unfamiliarised solution...

Education › Re: 2013/2014 Obafemi Awolowo University ::ASPIRANTS:: by Arithmetic(m): 12:39pm On Aug 10, 2014

Happy Sunday guys...

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 11:51pm On Aug 09, 2014

benbuks: guess others had done justice , any need for my solution.?

Sure boss.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 11:48pm On Aug 09, 2014

Kendzyma: rt(x^2+ax-1)+rt(x^2+bx-1)=rt(a)+rt(b)
rt(x^2+ax-1)=rt(a)
x^2+ax-1=a
x^2+a(x-1)=1..........eq 1
rt(x^2+bx-1)=rt(b)
x^2+bx-1=b
x^2+b(x-1)=1........eq 2
dividing eq 1 by eq 2
1+a/b=1
a=0
....subtituting a=0 in equation 1
x=+/-1
subtituting x=-1 in equation 2

b=0
subtituting x=1 in equation 2
b=0
hence b=a

You got one out of the two ans. x=1, but x is not equal to -1, try substituting into the ques. Infact u enlightened me, never thought of this method, should I give the 2nd ans. or you still want to try, you can help out 'cos I'm still working on the solution.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 1:04pm On Aug 08, 2014

benbuks , efficiencie , et al...

Arithmetic: 1)¥(x²+ax-1)+¥(x²+bx-1)=¥a+¥b.
2)¥(2^x+2+5)+¥(5.2^x-9)=¥(3.2^x+2+21).
Find x. NB:Let ¥ rep. square root.

benbuks , efficiencie , et al...

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 12:58pm On Aug 08, 2014

Drniyi4u: I got x = 2.3219
this is d working

Yeah, correct. x=log₂⁵. What about the first one.

Education › Re: 2013/2014 Obafemi Awolowo University ::ASPIRANTS:: by Arithmetic(m): 1:42pm On Aug 06, 2014

hausadreturn: oau is not on strike

I mean internal strike/shut down.

Education › Re: Nairaland Mathematics Clinic by Arithmetic(m): 1:31pm On Aug 06, 2014

1)¥(x²+ax-1)+¥(x²+bx-1)=¥a+¥b.
2)¥(2^x+2+5)+¥(5.2^x-9)=¥(3.2^x+2+21).
Find x. NB:Let ¥ rep. square root.

Education › Re: 2013/2014 Obafemi Awolowo University ::ASPIRANTS:: by Arithmetic(m): 3:24pm On Aug 05, 2014

Hello guys... This strike of a thing, may GOD help us...

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