Re: TI-M: Ooh...'nother integral question :)
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Re: TI-M: Ooh...'nother integral question :)
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)