What substitution would i use for

Wat how do you do this

do what? change of variable? or integrate?

I dont see why

(1- tan^2(x/2))/(1+tan^2(x/2)) = cosx

tan(x/2) = u
sec^2(x/2)*(1/2)dx = du
so, if your integral is in the form int[ R(sin(x),cos(x))]dx, then you can change it to
int[R(sin(x),cos(x)]*2cos^2(x/2)*sec^2(x/2)*(1/2)dx, which becomes int[R(((1-u^2)/(1+u^2)),(2u/(1+u^2)))]*2*(sqrt(1/(1+u^2)))du

[math]\frac{ 1 - \frac{sin^2(x/2) }{ cos^2(x/2) } }{1 + \frac{sin^2(x/2) }{ cos^2(x/2) } }[/math] is the same as
[math]\frac{ \frac{cos^2(x/2) - sin^2(x/2) }{ cos^2(x/2) } }{ \frac{cos^2(x/2) + sin^2(x/2) }{ cos^2(x/2) } }[/math]

if you simplify, you get [math] cos^2(x/2) - sin^2(x/2) [/math] which is [math] cos(x)[/math]

i fucked up the 2cos^2(x/2) calculation, the sqrt shouldn't be there/

Why not just set up the integral like

Sec(x)tan(x)sec^8(x)(sec^2(x)-1)

And then sub u=tanx. Come on Veeky Forums all these complicated methods when all u have to do is play around with trig identities and simple u substituion.

researchgate.net/publication/281255489_Rotor_Coordinates_Vector_Trigonometry_and_Kepler-Newton_Orbits
In case you want an intuition for what this "t" is.
(It's the half-turn in the paper.)

...

All trigonometric integrals goes like this:

2cos(x)=e^ix+e^-ix
2isin(x)=e^ix-e^-ix

It's from the rational parameterization of the unit circle. Do you even rational trigonometry. The t gives a point (0,t) on the y axis and the coordinate (cos(x),sin(x)) is where a line from (-1,0) through (0,t) intersects the unit circle. t=tan(x/2) is due to the inscribed angle theorem.

>i make things as difficult as possible so i can brag about it on Veeky Forums

This is all obvious for me, but how do I prove such statements?

oh shit, wrong thread

hint: norm.