₦airaland Forum

Master rhydex,i salute o

Integrate 7^sinx.dx

honey: take the subtitution e^x=v, then the que. bcums §v.15^v.dx..........dv/dx= e^x, and x= in.v, making dx= dv/v, replacing this in the integral §15^v.dv= 15^v+1/(v+1), replacing v with the original values of x, we have 15^(e^x+1)/(e^x+1)

...when you get to integral 15^udu...it's integrated to give (1/ln15)*15^u...PROOF:if y=§15^u...remember that e^ln15=15(from log)...substitute that...y=§e^uln15du....if a=uln15...da=ln15du...du=da/ln15...substitute those to give §e^a.da/ln15...(1/ln15)§e^ada...(1/ln15)e^a...but don't forget that a=uln15...(1/ln15)e^uln15...(1/ln15)e^ln15*u...don't forget that e^ln15=15...(1/ln15)15^u...it means §a^u=(1/lna)a^u

Welcom to the maths clinic...master,professor rhydex....let's keep the thread moving...

rhydex 247: integral xarcsinx dx.
solution
arcsinx implies sin@=x. drawing a right angle triangle with angle @. opp=x, hyp=1 and adj=sqrt(1-x^2). recalling from our SOHCAHTOA. Cos@=adj/hyp= sqrt(1-x^2). @= arcsinx. d@/dx= 1/ sqrt(1-x^2). dx=sqrt(1-x^2)d@. recall dat sqrt(1-x^2)=cos@. now substitute. we av
integral @sin@cos@d@
let u=@. du/d@=1 and v=sin@cos@. dv= sin^2@/2
using integration by part
uv- integral vdu
@sin^2@/2 - integral sin^2@/2 d@
@sin^2@/2 - 1/2 integral sin^2@d@.
@sin^2@/2 - 1/2 integral 1(1-cos2@)/2 d@
@sin^2@/2 - 1/4 [(integral d@ - integral cos2@d@)]
@sin^2@/2 - 1/4 (@-sin2@/2)
@sin^2@/2 - 1/4@+sin2@/8. now recall that sin2@=2sin@cos@
@sin^2@/2 -@/4 + 2sin@cos@/8
@sin^2@/2 - @/4 +sin@cos@/4
substituting we av
x^2arcsinx/2 - arcsinx/4 + xsqrt(1-x^2)/4
1/4[(2x^2-1)arcsinx+xsqrt(1-x^2)}+C

...that's my master,i salute o

lavosier: (1/2)x^2arcsinx-(1/4)arcsinx+1/4xsqreroot of (1-x^2)+c..ope am crrect??

...using integration by parts,u=arcsinx and dv=xdx...du=1/√1-x²...v=½x²...substitute that to the formula ∫udv=uv-∫vdu...by substituting to the r.h.s...½x²arcsinx-∫½x²/∫√1-x²...½x²arcsinx-½∫x²/√1-x²...EQUATION 1...to save space,taken the one with integral sign...½∫x²/√1-x².dx...if x=sin@...(i)...dx=cos@d@...½∫sin²@/√1-sin²@*cos@...½∫sin²@/√cos²@*cos@...½∫sin²@/cos@*cos@...½∫sin²@d@...{from trigonometry,cos2@=1-2sin²@...sin²@=(1-cos2@)/2}...½∫(1-cos2@)/2...¼∫1-cos2@...¼(@-sin2@/2)...{from trigonometry,sin2@=2sin@cos@}...¼(@-sin@cos@)...from...(i)...x=sin@...@=arcsinx...cos@=√1-x²...substitute those to get...¼(arcsinx-x√1-x²)...EQUATION 2(don't forget we solved ½∫x²/√1-x²).then substitute it to equation 1 above to give...½x²arcsinx-¼(arcsinx-x√1-x²)...½x²arcsinx-¼arcsinx+¼x√1-x²...

benbuks: Integrate(x arcsinx)dx..

...using integration by parts,u=arcsinx and dv=xdx...du=1/√1-x²...v=½x²...substitute that to the formula ∫udv=uv-∫vdu...by substituting to the r.h.s...½x²arcsinx-∫½x²/∫√1-x²...½x²arcsinx-½∫x²/√1-x²...EQUATION 1...to save space,taken the one with integral sign...½∫x²/√1-x².dx...if x=sin@...(i)...dx=cos@d@...½∫sin²@/√1-sin²@*cos@...½∫sin²@/√cos²@*cos@...½∫sin²@/cos@*cos@...½∫sin²@d@...{from trigonometry,cos2@=1-2sin²@...sin²@=(1-cos2@)/2}...½∫(1-cos2@)/2...¼∫1-cos2@...¼(@-sin2@/2)...{from trigonometry,sin2@=2sin@cos@}...¼(@-sin@cos@)...from...(i)...x=sin@...@=arcsinx...cos@=√1-x²...substitute those to get...¼(arcsinx-x√1-x²)...EQUATION 2(don't forget we solved ½∫x²/√1-x²).then substitute it to equation 1 above to give...½x²arcsinx-¼(arcsinx-x√1-x²)...½x²arcsinx-¼arcsinx+¼x√1-x²...

Integral e^(x^2)

benbuks: ..show working.

...remember that from partial differentiation...dz/dx=§z/§x+§z/dy*dy/dx where § is the partial differentiation...from the equation...arctan(x/y)=y/x...multiply both sides by tan...tan.arctan(x/y)=tan(y/x)...x/y=tan(y/x)...x/y-tan(y/x)=0...let z=0...x/y-tan(y/x)=z...from partial differentiation...§z/§x=1/y+1/x^2.sec^2(y/x)...§z/§y=-x/y^2-1/xsec^2(y/x)...dz/dx=0...substitute all to the formula...dz/dx=§z/§x+§z/§y*dy/dx...and make dy/dx the subject of formula...

benbuks: Y/x =arctan (x/y)....compute its differential coefficient.

...dy/dx=y/x...

I'm really happy,that some generals have come to the thread...kudos to the everybody here...

Pls,explain ur answer bro@lavosier.

Ortarico: Ok, i've corrected myself by making dv = x, u = arcsinx
Got the answer to be: (x^2)arcsinx + |_/1-x^2)|^1/2 + c

...my oga at the top...u will still integrate the square root of (1-x^2)...

lavosier: (1/2)x^2arcsinx-(1/4)arcsinx+1/4xsqreroot of (1-x^2)+c..ope am crrect??

...maybe...but pls,try to solve any questions you can solve my great bro...we are here not to test anyone's knowledge but to brainstorm and help one another

Ortarico: Hmmmmmm. . . .

§ xarcsinx
integrating by parts:
let u = x; dv = arcsinx
du = (1)dx; v = 1/_/1 - x^2
du = dx; v = 1/_/1 - x^2

General formula of integrating by parts: uv - § vdu
x.1/_/(1-x^2) - § 1/_/(1-x^2)

using the u-substitution for § 1/_/(1-x^2):
u = 1-x^2
du = -2x dx
-1/2du = dx

x/_/(1-x^2) - § 1/_/u.(-1/2) + c
factorise out the -1/2:
x/_/(1-x^2) + 1/2 § u^(-1/2) + c
x/_/(1-x^2) + 1/2 § u^1/2/1/2 + c
x/_/(1-x^2) + 1/2.u^1/2/1/2 + c
x/_/(1-x^2) + |x|^1/2 + c

...you have done a great job my brother...but where you put dv=arcsinx...v will be integration of arcsinx...which makes the question complex

.i support that general...ISAAC NEWTON is my greatest master...TO DO WHAT PEOPLE HAVE NOT DONE...YOU HAVE TO START FROM WHAT PEOPLE HAVE DONE...

Richiez: welcome bro! you definitely will be great
I remember the words of Sir Isaac Newton "If I have seen further than others, it is by standing upon the shoulders of giants"
ISAAC NEWTON WAS A GREAT MATHEMATICIAN BECAUSE HE LEARNT FROM OTHERS AND ADDED MORE TO WHAT HE LEARNT

...i support that general...ISAAC NEWTON is my greatest master...TO DO WHAT PEOPLE HAVE NOT DONE...YOU HAVE TO START FROM WHAT PEOPLE HAVE DONE...

Ortarico: Spot on bruv. Ehen, you said that the no 2 of your questions you posted at the gurus' thread has been solved. Would you mind sharing the solution with me pls?

...i'm not that very good at hyperbola...but using the equation in canonical form...(x-x')²/a²-(y-y')²/b²=1...remember that the asymptotes of an hyperbola is ±b/a and the asymptotes of the question is ±3...it means b=3 and a=1...which passes through the points 4,6...substitute that to the question to get...(x-4)²/1²-(y-6)²/3²=1...(x-4)²-(y-6)²/9=1...then complete it sir...

Sir Chides: i think datz x/root(1-x^2) -root(1-x^2)+c

...hmmm,no sir...try more sir...let me give you hint...use integration by parts...let u be arcsinx and v be x...

....Integrate xarcsinx.dx....

Morning to all my generals...

honey: general, i am awaiting promotion from leiutenant to general

...i respect your mathematical approach bro...me not yet a maths corporal let alone lieutenant.. .

Uniben...about to study mechanical engineering

My generals are here...that's great...let me take a seat and watch as my ogas they perform their magic here

Thank you very much...generals richiez and double dx...thanx for the compliments but the best way to be great on this earth is to know that one knows nothing...for we young mathematicians,we need God most and people like you to help us...PLS DON'T FAR AWAY FROM US...a tree doesn't make a forest...let's keep the thread moving...THANK YOU VERY MUCH SIR...

Kudos to everybody here

[quote author=t.boy007](math):
Factorize:
x^4 - 8
(physics):
when an ebonite is rubbed with fur,what will be the charge of the two substances respectively?
(chemistry):
what is the substance used to test for ALKANOLS?[/quote]...x^4-8...x^4-8-8+8...x^4-16+8...(x^2)^2-4^2+8...since a^2-b^2=(a+b)(a-b)...therefore...(x^2+4)(x^2-4)+8...since x^2-4 is also a difference of two squares...(x^2+4)(x^2-2^2)+8...(x^2+4)(x+2)(x-2)+8.....(2)...when an ebonite rod is rubbed with fur...the ebonite is negatively charged while the fur is positively charged coz electrons are transferred from the atoms of the fur to the atoms of the ebonite...since the ebonite gain an electron,then it becomes negatively charged and fur which lose electron will become positively charged...(3)...hmmm.it can be differentiated from others by using lucas reagent

Mr Calculus: my bro i greet u powerfully sir

...my oga i salute you...

Miscellaneous: Calculusf(x) i see u o. . . . . My main baddo. . .the only oga

...i dey low key ooo my oga

You guys are doing great here ooo...let me sit and wait my ogas as they are doing their magic

Let's bring tough questions pals,and let's gain from one another.....here is the first one i have with me here....solve for x,if 2^x=5x+11

lavosier: Maro festus okezzi..

...that's great my brothers...let's keep the thread moving...thanx to my main men...ortarico and mr calculus...more power to your elbow