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Arithmetic's Posts

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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 / {3x + 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 / {3x + 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.

1 Like 1 Share

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=729a-1,
=>32.2a=36(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)¥(2x+2+5)+¥(5.2x-9)=¥(3.2x+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)¥(2x+2+5)+¥(5.2x-9)=¥(3.2x+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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