Re: TI-M: Ooh...'nother integral question :)
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Re: TI-M: Ooh...'nother integral question :)
In a message dated 9/24/00 2:47:48 PM Mountain Daylight Time,
JasonScho@aol.com writes:
> sin(x)^3/(sin(x)^3+cos(x)^3) = tan(x)^3/(tan(x)^3+1)
>
> Let u=tan(x)
> du=sec(x)²dx=(1+u²)dx
>
> fnInt(sin(x)^3/(sin(x)^3+cos(x)^3),x) = fnInt(u^3/((u+1)(u²-u+1)(u²+1)),u)
> =
> fnInt((1/6)(-1/(u+1)+(4u-2)/(u²-u+1)+(3-3u)/(u²+1)),u) by partial fraction
> decomposition
> =
> 1/6*(-ln(u+1)+fnInt(2(2u-1)/(u²-u+1)+(-1.5(2u)+3)/(u²+1),u)
> = -(ln(u+1))/6 +
> (ln(u²-u+1))/3 - (ln(u²+1))/4+arctan(u)/2
> = -(ln(tan(x)+1))/6 +
> (ln(sec(x)^2-tan(x)))/3 - (ln(sec(x))/2 + x/2
>
> This method can be used for later values of n
>
> u^4+1=(u²+u*sqrt(2)+1)(u²-u*sqrt(2)+1)
> u^5+1=(u+1)(2u²-(1+sqrt(5))u+1)(2u²-(1-sqrt(5))u+1)/4
> u^6+1=(u²+1)(u²+u*sqrt(3)+1)(u²-u*sqrt(3)+1) hard due to doubled (u²+1)
> factor
> u^7+1=beats me (good luck)
Ahhh...I see, thanks :)
JayEll