NewStats: 3,263,866 , 8,181,654 topics. Date: Sunday, 08 June 2025 at 11:22 AM 2r6525

if a<b<0 and a^2+b^2=4ab, evaluate a+b/a-b.

from benbuks question integral [sqrt(1+(sqrtx)]/x dx
SOLUTIION.
FOR THE INTEGRAND [SQRT(SQRTX)+1]/Xdx
by substitution method
let u= sqrt(x) , x=u^2
du/dx=1/2sqrt(x)..... dx=2sqrt(x)du
integral [sqrt(u+1)*2sqrt(x) ]du/u^2
recall that sqrt(x)=u
integral [sqrt(u+1)*2u] du/u^2
2 integral sqrt(u+1)du/u
For the integral sqrt(u+1)du/u.... substitute s=sqrt(u+1)... ds/du= 1/2sqrt(u+1).....du= 2sqrt(u+1)ds...... from s=sqrt(u+1).... s^2=u+1.....s^2-1=u
2 integral s(2s)ds/s^2-1........2 integral2s^2ds/s^2-1.......4 integral s^2ds/s^2-1......
For the integrand s^2/s^2-1, do long division we have this
4 integral [1+1/2(s-1)-1/2(s+1)]ds
4 [integral ds +1/2 integral ds/(s-1) -1/2 integral ds/(s+2)]
4s+2ln(s-1)-2ln(s+1)+C
put back s = sqrt(u+1). we have
(4sqrt(u+1))+(2lnsqrt(u+1)-1) - (2lnsqrt(u+1)+1)+C
substitute u=sqrt(x)
we have
4sqrt[sqrt(x)+1]+2lnsqrt[sqrt(x)+1]-1 -2lnsqrt[sqrt(x)+1]+1 +C
which is equivalent to
4sqrt[sqrt(x)+1] -4tanh^-1[sqrt[sqrt(x)+1]. all is well.

@ d citizen.
From y=x^x^x. dy/dx=x^x^x[x^x(lnx+1)lnx+x^x/x]. Now y=x^(x^x^x).
take log of both sidés.
lny=lnx^(x^x^x).
lny=x^x^xlnx.
d/dx(lny)=d/dx(x^x^xlnx).
dy/dx(1/y)=udv/dx+vdu/dx. Take u=x^x^x du/dx=x^x^x[x^x(lnx+1)lnx+x^x/x]. And v =lnx dv/dx=1/x.
dy/dx=y[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x].
now recall that y=x^x^x^x. Therefore dy/dx=x^x^x^x[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x].
similarly for y=x^(x^x^x^x) Take log of both sides.
lny=x^x^x^xlnx
d/dx(lny)=d/dx(x^x^x^xlnx). Take u=x^x^x^x du/dx=x^x^x^x[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x]. And take v=lnx dv/dx=1/x . Using product rule. We av
(1/y)dy/dx=x^x^x^x/x+lnx(x^x^x^x[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x].
dy/dx=y[x^x^x^x/x+lnx(x^x^x^x[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x]]. But y=x^x^x^x^x. Therefore dy/dx=x^x^x^x^x([x^x^x^x/x+lnx(x^x^x^x[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x]]). For easier understanding let y=X^x^x^x^x.
u=x^x^x^x.
v=x^x^x.
w=x^x. Hence
dy/dx=yu/x+uvy[(w(lnx+1)lnx]+w/x+w/x. All is well.

@ d citizen
i have the solution to ur questiön 3^x+4^x=5^x. Buh is a graphical solution which indicate dat x=2. I dnt knw hw to paste it ön nairaland.

@ d citizen
hint on how to solve y=x^x^x^x^x.
soln.
let me start 4rm y=x^x. Take log of both sides.
lny=xlnx. d/dx(lny)=d/dx(xlnx)
1/y(dy/dx)=lnx+1.
dy/dx=y(lnx+1). Recall that y=x^x.
dy/dx=x^x(lnx+1).
Similarly for y=x^(x^x).
Take log of both sides. We av
lny=lnx^(x^x). Which implies.
lny=x^xlnx. d/dx(lny)=d/dx(x^xlnx).
dy/dx(1/y)= d/dx(x^x)*lnx+d/dx(lnx)x^x.
now recall that d/dx(x^x)=x^x(lnx+1) and d/dx(lnx)=1/x.
dy/dx(1/y)=x^x(lnx+1)lnx+1/x*x^x.
dy/dx=y[x^x(lnx+1)lnx+x^-1+x]. Recall that y=x^x^x. Hence we av dy/dx=x^x^x[x^x(lnx+1)lnx+x^(x-1)].
use this method for y=x^(x^x^x) and y=x^(x^x^x^x). All is well.

d/dx(x^x^x^x^x)=x^x^x^x^x+x^x^x[x^x^xlogx(x^xlogx(1/x+log^2x+logx)+1/x)+1/x].

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Derivative:

d\/dx(x^(x^(x^(x^x)))) = x^(x^(x^(x^x))) (x^(x^(x^x)-1)+x^(x^(x^x)) log(x) (x^(x^x-1)+x^(x^x) log(x) (x^(x-1)+x^x log(x) (log(x)+1)))

dx(x^(x^(x^(x^x))))
Express x^(x^(x^(x^x))) as a power of e: x^(x^(x^(x^x))) = e^(log(x^(x^(x^(x^x))))) = e^(x^(x^(x^x)) log(x))
= d\/dx(e^(x^(x^(x^x)) log(x)))
Using the chain rule, d\/dx(e^(x^(x^(x^x)) log(x))) = de^u/du* du\dx, where u = x^(x^(x^x)) log(x) and d\du(e^u) = e^u = d\dx(x^(x^(x^x)) log(x))) e^(x^(x^(x^x)) log(x))
Express e^(x^(x^(x^x)) log(x)) as a power of x: e^(x^(x^(x^x)) log(x)) = e^(log(x^(x^(x^(x^x))))) = x^(x^(x^(x^x)))
= x^(x^(x^(x^x))) d\/dx(x^(x^(x^x)) log(x))
Use the product rule, d/dx(u,v) = v du/dx+udv)/dx, where u = x^(x^(x^x)) and v = log(x)
= log(x) d\/dx(x^(x^(x^x)))+x^(x^(x^x)) d\/dx(log(x)) x^(x^(x^(x^x)))
Express x^(x^(x^x)) as a power of e: x^(x^(x^x)) = e^(log(x^(x^(x^x)))) = e^(x^(x^x) log(x))
= x^(x^(x^(x^x))) (x^(x^(x^x)) (d/dx(log(x)))+d/dx(e^(x^(x^x) log(x))) log(x))
Using the chain rule, d/dx(e^(x^(x^x) log(x))) =de^u/du* du\dx,where u = x^(x^x) log(x) and d/du(e^u) = e^u
= x^(x^(x^(x^x))) (x^(x^(x^x)) (d/dx(log(x)))+d/dx(x^(x^x) log(x)) e^(x^(x^x) log(x)) log(x))
Express e^(x^(x^x) log(x)) as a power of x: e^(x^(x^x) log(x)) = e^(log(x^(x^(x^x)))) = x^(x^(x^x))
= x^(x^(x^(x^x))) (x^(x^(x^x)) d/dx(x^(x^x) log(x)) log(x)+x^(x^(x^x)) (d/dx(log(x))))
Use the product rule, d/dx(u, v) = v ( du)/( dx)+u ( dv)/( dx), where u = x^(x^x) and v = log(x)
= x^(x^(x^(x^x))) (x^(x^(x^x)) (d\/dx(log(x)))+log(x) d\/dx(x^(x^x))+x^(x^x) d\/dx(log(x)) x^(x^(x^x)) log(x))
Express x^(x^x) as a power of e: x^(x^x) = e^(log(x^(x^x))) = e^(x^x log(x))
= x^(x^(x^(x^x))) (x^(x^(x^x)) (d/dx(log(x)))+x^(x^(x^x)) log(x) (x^(x^x) (d/dx(log(x)))+d/dx(e^(x^x log(x))) log(x)))
The derivative of log(x) is 1/x
= x^(x^(x^(x^x))) (x^(x^(x^x)) log(x) (x^(x^x) (d/dx(log(x)))+log(x) (d/dx(e^(x^x log(x)))))+1/x x^(x^(x^x)))
Simplify the expression gives
= x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(x^x) (d/dx(log(x)))+log(x) (d/dx(e^(x^x log(x))))))\nThe derivative of log(x) is 1\/x:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (log(x) (d\/dx(e^(x^x log(x))))+1\/x x^(x^x)))\nSimplify the expression:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+log(x) (d\/dx(e^(x^x log(x))))))\nUsing the chain rule, d\/dx(e^(x^x log(x))) = ( de^u)\/( du) ( du)\/( dx), where u = x^x log(x) and ( d)\/( du)(e^u) = e^u:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+d\/dx(x^x log(x)) e^(x^x log(x)) log(x)))\nExpress e^(x^x log(x)) as a power of x: e^(x^x log(x)) = e^(log(x^(x^x))) = x^(x^x):\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) d\/dx(x^x log(x)) log(x)))\nUse the product rule, d\/dx(u v) = v ( du)\/( dx)+u ( dv)\/( dx), where u = x^x and v = log(x):\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+log(x) d\/dx(x^x)+x^x d\/dx(log(x)) x^(x^x) log(x)))\nExpress x^x as a power of e: x^x = e^(log(x^x)) = e^(x log(x)):\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x (d\/dx(log(x)))+d\/dx(e^(x log(x))) log(x))))\nUsing the chain rule, d\/dx(e^(x log(x))) = ( de^u)\/( du) ( du)\/( dx), where u = x log(x) and ( d)\/( du)(e^u) = e^u:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x (d\/dx(log(x)))+d\/dx(x log(x)) e^(x log(x)) log(x))))\nExpress e^(x log(x)) as a power of x: e^(x log(x)) = e^(log(x^x)) = x^x:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x d\/dx(x log(x)) log(x)+x^x (d\/dx(log(x))))))\nUse the product rule, d\/dx(u v) = v ( du)\/( dx)+u ( dv)\/( dx), where u = x and v = log(x):\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x (d\/dx(log(x)))+log(x) d\/dx(x)+x d\/dx(log(x)) x^x log(x))))\nThe derivative of x is 1:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x (d\/dx(log(x)))+x^x log(x) (x (d\/dx(log(x)))+1 log(x)))))\nThe derivative of log(x) is 1\/x:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x log(x) (log(x)+x (d\/dx(log(x))))+1\/x x^x)))\n
Simplify the expression= x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^(-1+x)+x^x log(x) (log(x)+x (d\/dx(log(x)))))))\nThe derivative of log(x) is 1\/x:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^(-1+x)+x^x log(x) (log(x)+1\/x x))))\n
Simplify the expression:
Answer: = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^(-1+x)+x^x log(x) (1+log(x)))))
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x + y = 5.......(1)
x^x + y^y =31.......(2)

from eqn(1), x=5-y, hence we substitute this value for x in eqn(2).
(5-y)^(5-y) + y^y = 31.......(3)
now, we have reduced the problem to what value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3
i.e (5-3)^(5-3) + 3^3
2^2 + 27
4+27=31
now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x
recall that x+y=5
therefore; x+3=5 or x=5-3=2
hence, x=2 when y=3. therefore we av (2,3)

[color=#990000][/color]

prove that 1=2
solution
1 can never be equal to 2. hence the answer is FALSE.
[color=#770077][/color]

integral e^tanxdx
solution
using wolfram mathematical online integrator
-1/2ie^-id[(e^2iEi(tanx-i)-Ei(tanx+i)].

I don't think there's an analytical solution. for integral e^tanxdx

If you try it with Wolfram's online integrator, it gives the result in of the Exponential Integral Ei function, which I believe is transcendental.

QUESTION F(X)=3/X^3. Using ist principle method
SOLUTION
Let y=3/x^3
i will let my change in x =h and change in y =k
y+k=x+h
y+k=3/(x+h)^3
y+k=3/x^3+3x^2h+3xk^2+k^3
subtract y from both sides
y+k-y=3/x^3+3x^2h+3xh^2+h^3 -y
but y=3/x^3
k=3/x^3+3x^2h+3xh^2+h^3 -3/x^3
taking the l.c.m
k=3x^3 - 3(x^3+3x^2h+3xh^2+h^3) /x^3(x^3+3x^2h+3xh^2+h^3)
k=3x^3 - 3x^3-9x^2h-9xh^2-3h^3 /x^3(x^3+3x^2h+3xh^2+h^3)
k=-9x^2h-9xh^2-3k^3/x^3(x^3+3x^2h+3xh^2+h^3)
divide through by change in x=h
k/h= -9x^2h-9xh^2-3h^3/x^3(x^3+3x^2h+3xh^2+h^3)*1/h
k/h= -9x^2-9xh-3h^2/x^3(x^3+3x^2h+3xh^2+h^3)
k/h=dy/dx=lim h---->0[-9x^2-9xh-3h^2/x^3(x^3+3x^2h+3xh^2+h^3)]
dy/dx=-9x^2/x^3(x^3)
dy/dx=-9x^2/x^6=-9/x^4
therefore dy/dx=-9x^-4. good luck


QUESTION Lim--->x=pi(sin(picosx)/(x-pi)^2)
solution
we all know that pi=180
putx=pi=180
we obtain 0/0. which is indeterminate.
using l'hopitals rule which says
limx--->a (f'(x)/g'(x))
when you differentiate sin(picosx) w.r.t.
i.e f'(x)=cos(picosx)(-pisinx)
g'(x)= 2(x-pi)
limx--->pi[cos(picosx)(-pisinx)/2(x-pi)]
when we put x=pi we obtain 0/0 which is indeterminate
we will differentiate again untill we obtain a real value
f''(x)=cos(picosx)(-picosx)+(-pisinx)(sin(picosx))(pisinx)
g''(x)= 2
limx---->pi[(cos(picosx)(-picosx)+(-pisinx)(sin(picosx)(pisinx)/2]
putx=pi=180
we obtain [(-1)(180)+0/2]
which implies -90 or -pi/2.

1. find the orthogonal trajectory of the curve y^3+3x^2y=c1.

the real numbers a,b,c satisfy a nt equal to b and 2009(a-b)+ sqrt 2009(b-c)+(c-a)=0. Find the value of (c-b)(c-a)/(a-b)^2.
Solution.
Let x= sqrt 2009. Also x^2= 2009.
puttin it in the eqn gives x^2(a-b+x(b-c)+(c-a)=0.
a nt equal to b => this eqn is a quadratic eqn. Obviously x=sqrt 2009 and 1 are the two root of this quadratic eqn. Nw recall dat.
sum of x =-b/a
product of root =c/a. Sum of x.
=> sqrt 2009 +1=c-b/a-b. Also
product of root
=> sqrt 2009*1= c-a/a-b. Hence (c-b)(c-a)/(a-b)(a-b).
=>(sqrt 2009 +1)(sqrt 2009)
=2009+sqrt 2009.

well done my oga, u are great. We av two ways to solve d questn. Ur method is ok. Here is another way. let y=x+1, x=y-1. Put x=y-1 in d eqn gives (y-2)^4+(y+2)^4.
(y^2-4y+4)^2+(y^2+4y+4)^2=82. Reduces to
y^4+24y^2-25=0
(y^2+25)(y^2-1)=0.
since y^2+25>0 then y^2-1=0. So y=1 or y=-1. But recall that x=y-1. Hence x=0 or x=-2.
We are going to take mathematics to another level.

1. Solve the equation. (x-1)^4+(x+3)^4=82.
2. The real numbers a,b,c satisfy a<>b and 2009[a-b]+sqrt(2009)[b-c]+[c-a]=0. Find the value of (c-b)(c-a)/(a-b)^2.
3. Solve the systems of eqn.
x^2+xy+y^2=84...eqn 1 and x+sqrt(xy)+y=14...eqn 2.

3). Expand f(x,y)=x^2e^y at the point (1,0) up to in second degree.
solution to question 3.
we are going to some language here says
f subscript x=> fx
f subscript y=> fy.
Let h=change in x => h=x-xo. And
Let k=change in y => k= y-yo.
(1,0)=> xo=1 and yo=0. h=(x-1) and k=y-0=y. i.e k=y.
f(x,y)=x^2e^y at f(1,0)= 1.
fx=2xe^y at f(1,0)=2.
fxx=2e^y at f(1,0)=2.
fxy=2xe^y at f(1,0)=2.
fy=x^2e^y at f(1,0)=1.
fyy=x^2e^y at f(1,0)=1.
Now recall the formula of fuctions of two variables (taylor series nd maclaurin series).
Pls take Note of this 4mula.
f(x,y)= f(xo,yo)+hfx(xo,yo)+kfy(xo,yo)+1/2![h^2fxx(xo,yo)+2hkfxy(xo,yo)+k^2fyy(xo,yo)]+1/3![h^3fxxx(xo,yo)+3h^2kfxxy(xo,yo)+3k^2hfyyx(xo,yo)+k^3fyyy(xo,yo)]. From the questn we are asked to stop at in second degree. So we neglect the third degree. By putting al d values we av. Dnt 4get dat h=x-1 and k=y.
f(x,y)=1+(x-1)(2)+1/2![(x-1)^2(2)+2(x-1)(y)(2)+y^2(1)]. Try to simplify dis. We ave.
f(x,y)=x^2-y+2xy+v^2/2.
Quote: Always take the high road.

2). Solve (8y-x^2y)dy/dx+(x-xy^2)=0. Hint: exact eqn.
solution to questn 2.
Q(x,y)dy/dx+P(x,y)=0...... Q(x,y)dy+P(x,y)dx=0. For the eqn to be exact it must satisfy dis
dP(x,y)/dy=dQ(x,y)/dx.
P(x,y)=du/dx
Q(x,y)=du/dy.
Q(x,y)=8y-x^2y
P(x,y)=x-xy^2.
dQ(x,y)/dx=-2xy
dP(x,y)/dy=-2xy. Hence the eqn is exact.
du/dx=P(x,y)=x-xy^2...... du/dx=x-xy^2.
integratin both sides w.r.t x
U=x^2/2-x^2y^2/2+T(y). Where T(y) is an arbitrary function of y.
differentiating U w.r.t. y
du/dy=-x^2y+T'(y).
du/dy=Q(x,y)=8y-x^2y.
-x^2y+T'(y)=8y-x^2y.
T'(y)=8y. Integrate w.r.t y.
T(y)=4y^2+C. But U=x^2/2-x^2y^2/2+T(y).
U(x,y)=x^2/2-x^2y^2/2+4y^2+C. OR
U(x,y)=x^2(1-y^2)/2 +4y^2+C.
check dis @ honey kip it up.

1). solve. (3xy+3y-4)dx+(x+1)^2dy=0.
solution to Question 1.
(x+1)^2dy=-(3xy+3y-4)dx..... dy/dx=-(3xy+3y-4)/(x+1)^2..... dy/dx+3xy+3y-4/(x+1)^2=0..... dy/dx+3xy/(x+1)^2+3y/(x+1)^2-4/(x+1)^2=0....
dy/dx+3xy+3y/(x+1)^2 =4/(x+1)^2.
dy/dx+3y(x+1)/(x+1)^2=4/(x+1)^2.
dy/dx+3y/(x+1)=4/(x+1)^2.
Now recall that dy/dx+P(x)y=Q(x). Which is called Linear equation Or Integrating Factor (I.F).
P(x)=3/(x+1) and Q(x)=4/(x+1)^2.
I.F= e^§P(x)dx. Where § rep. integral symbol.
I.F= e^§3dx/(x+1)
I.F= e^3ln(x+1) = (x+1)^3.
I.F= (x+1)^3.
y(I.F)= §Q(I.F)dx
y(x+1)^3= § 4/(x+1)^2*(x+1)^3 dx.
y(x+1)^3= 4§(x+1)dx.
y(x+1)^3=4[(x+1)^2/2]+C.
y(x+1)^3=2[(x+1)^2]+C.
divide both sides by (x+1)^3.
y=2/(x+1) +C/(x+1)^3. OR
y=2(x+1)^-1 + C(x+1)^-3. Hmmm all is well. Check dat @ honey. U tried.

Dis questions goes to my fellow mathematician.(Break fast)
1). Solve (3xy+3y-4)dx+(x+1)^2dy=0.
2). Solve (8y-x^2y)dy/dx+(x-xy^2)=0.hint: exact equation
3). Expand f(x,y)=x^2e^y at the point (1,0) up to in second degree.

integral sqrt 5x^2 dx.
soln.
We all know that sqrt of 5x^2 =>(sqrt5)x. By putting it back in d questn. We av integral (sqrt5)x dx.
sqrt5(x^2/2).

integral x^2/x^3+5 dx.
Soln.
let u=x^3+5. du/dx=3x^2. dx=du/3x^2. By substitution we ave. Integral x^2/u*du/3x^2.
1/3 integral 1/udu
1/3(lnu)+C. Recall that u=x^3+5.
1/3[ln(x^3+5)]+C.

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