Quick Question (1 Viewer)

[mazza_728]

How do u integrated: sqrt(sinxcos³x)??

 

[mervvyn]

try "t" method --> t = tan(x/2)

 

[ngai]

1/8*(4*cos(x)^5-8*cos(x)^3+((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(x)^3+((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(x)^3+((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(x)*I-I*((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(x)^2+((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*I+((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(x)^2-I*((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(x)+((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(x)^2+4*cos(x)-((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(x)-((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(x)+((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(x)^3*I+((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(x)^2*I-I*((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(x)^3-((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))-I*((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))-((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2)*EllipticPi(((1-cos(x)+sin(x))/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2)))*(sin(x)*cos(x)^3)^(1/2)/cos(x)^2/sin(x)^3

 

[Archman]

and all that by inspection.

 

[CrashOveride]

Ngai is only joking. Mazza just use the "t" method and simplify it works out, really.

 

[McLake]

"ElipticPi"? Expalin that to us ngai ...

 

[BillyMak]

My head hurts.

 

[ressul]

Any solution?

I found it's very complex to work it out by using t=tanx/2.

 

[Xayma]

"ElipticPi"? Expalin that to us ngai ...

It has something to do with an integral of φ.

Or from wolfram it is the elliptic integral of the third kind:

http://mathworld.wolfram.com/EllipticIntegraloftheThirdKind.html

Where Im guessing the k value is the one after the ellipticpi above.

 

[mojako]

hohoho..
I drew this by inspection...
using coloured pens

it's called "arts"
and should fall under "Appreciating the Beauty and Elegance (extracurricular topics)"

in beauty.GIF
blue is the graph of sqrt(sinxcos^3x)
red is its derivative
green is the graph of sqrt(sinxcos3x)
purple is its derivative

in beauty2.GIF
is the graph of 1/8*(4*cos(x)^5-8*cos(x)^3+((1-cos(x)+sin(x))/sin(x))^(1/2)*2^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*((-1+cos(x))/sin(x))^(1/2))
another quality release of mojako's GrApHiCs DESign sOCiETy

 

[Veck]

forget t results

S = integral sign

S root(sinxcos^3x)

= S root(sinx(1-sin^2 x)cosx)
= S root(sinxcosx - sin^3 x.cosx)
= S root(sinx (cosx - sin^2 x.cosx))
= S root(sinx (cosx - (1 - cos^2 x) cosx)
= S root(sinx (cosx - cosx + cos^2 x))
= S root(sinx(cos^2 x)
= S cosx root(sinx)
= 2/3 sin^3/2 x + C
= 2/3.sinx.root(sinx) + C

hmm.. how's that?

 

[gordo]

please assure me we arn;t going to asked that

 

[Archman]

forget t results

S = integral sign

S root(sinxcos^3x)

= S root(sinx(1-sin^2 x)cosx)
= S root(sinxcosx - sin^3 x.cosx)
= S root(sinx (cosx - sin^2 x.cosx))
= S root(sinx (cosx - (1 - cos^2 x) cosx)
= S root(sinx (cosx - cosx + cos^2 x))
= S root(sinx(cos^2 x)
= S cosx root(sinx)
= 2/3 sin^3/2 x + C
= 2/3.sinx.root(sinx) + C

hmm.. how's that?

i reckon something is wrong in that:
S root(sinxcos^3x)
.
.
.
= S root(sinx(cos^2 x)

 

[mojako]

i reckon something is wrong in that:
S root(sinxcos^3x)
.
.
.
= S root(sinx(cos^2 x)

no, its called a distortion of algerba and occurs naturally after repeated algebraic manipulation

 

[Veck]

hmm.. I thought so too.. then I checked it again and I can't see anything wrong with it at all

the line between these two

S root(sinxcos^3x)

= S root(sinx(1-sin^2 x)cosx)

is S root(sinx.cosx.cos^2 x)

which is legit...?hmmm...

EDIT: Sorry, check that, I didn't quite catch on to what you were saying. I'll look again

 

[Veck]

= S root(sinx (cosx - sin^2 x.cosx))
= S root(sinx (cosx - (1 - cos^2 x) cosx)
= S root(sinx (cosx - cosx + cos^2 x))

shit shit shit and damn


that should be S root sinx (cosx - cosx + cos^3 x)

which of course brings us back to where we started

 

[CrashOveride]

its natural algebra decay

 

[mojako]

its natural algebra decay

true true...
radioactive algebraic decay, it is.
half life of 30 minutes.
now it seems to have disappeared almost completely... approaches zero.