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The Weierstrass t substitution, also called the Universal Trig Substitution, is t=tan(x/2).  In order for this substitution to be useful for solving integrals involving such things as cos(x), tan(x), etc, it is necessary to express these trig functions in terms of tan(x/2), and then make the substitution.

formula Weierstrass
tan(x) = 2 tan(x/2)/(1-tan2(x/2)) tan(x)
   = 2t/(1-t2)
from the tan-of-sum formula, using tan(x/2 + x/2)
cos(x) = (1-tan2(x/2))/(1+tan2(x/2)) cos(x)
   = (1-t2)/(1+t2)
sin(x) = cos(x)tan(x)
     = 2 tan(x/2)/(1+tan2(x/2))
   = 2t/(1+t2)
by combining the two formulas above.
dx = 2 cos2(x/2) dt
    = (cos(x)+1) dt
dx = ((1-t2)/(1+t2) + 1) dt
   = 2dt/(1+t2)
t = tan(x/2), so dt/dx = (1/2) sec2(x/2), or
x=2 arctan(t), so dx/dt = 2/(t2+1)


The problem is to find the area of the region given by

x^4+y^4 ≤ x^2-x^2y^2+y^2

Rearranging, the boundary is the solution other than (0,0) of


So  factor the LHS as

(x^2+y^2-xy)(x^2+y^2+xy)=x^2+y^2, then convert to polar as

 . . . . . . format this a little better. 

(r^2-r^2cos t sin t)(r^2 + r^2 cos t sin t)=r^2, then factor out r^2
r^2(1-cos t sin t)(1+cos t sin t)=1
r^2 = -4/(sin(2t)^2-4)
r^2 = 8/(8-2sin(2t)^2)
r^2 = 8/(8+cos(4t)-1)
r^2 = 8/(cos(4t)+7)

Solve x^4+y^4=x^2-x^2y^2+y^2 for y
y=(1/sqrt(2)) (sqrt(-x^2+1+sqrt(-3x^4+2x^2+1))), with both sqrt's being plus or minus

Mathematica can't integrate this.

r^2 = 8/(cos(4t)+7)
area of wedge of angle dt is (1/2) r^2 dt, which is 4/(cos(4t)+7)
The integral of 4/(cos(4t)+7)dt is (1/sqrt(12)) arctan(sqrt(3)/2 tan(2t)), which eval at 0,2pi is 2pi/sqrt(3)

integral of 4/(cos(4x)+7)dx using Weierstrass t-substitutions, where t=tan(2x), and dt=2sec(2x)^2dx, so
cos(2x)^2 = (1+cos(4x))/2
cos(2x)^2 = (1+(1-t^2)/(1+t^2))/2 = (1+t^2+1-t^2)/(1+t^2)/2 = 1/(1+t^2)



integral is arctan(sqrt(3)t/2)/(2sqrt(3))

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Table of Integrals

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