solve the equation Integral sin^2xdx+Integral cos^2xdx=(x^3-4x)/(x-2)

We'll re-write the etrms from the left side of the
equation, solving the integrals.

We'll use the half angle
identities:

[cos (a/2)]^2 = (1 + cos
a)/2

[sin (a/2)]^2 = (1 - cos
a)/2

Int (sin x)^2 dx = (1/2)Int dx - (1/2)Int cos 2x dx =
x/2 - sin 2x/4 +c

Int (cos x)^2 dx  = x/2 + sin 2x/4 +
c

Now, we'll solve the equation substituting the integrals
by their results:

x/2 - sin 2x/4 + x/2 + sin 2x/4 = 
(x^3-4x)/(x-2)  

We'll eliminate like terms and we'll
factorize by x to the right side:

2x/2 =
x(x^2-4)/(x-2)

We'll re-write the difference of squares
from numerator: x^2 - 4 = (x-2)(x+2)

2x/2 =
x(x-2)(x+2)/(x-2)

We'll simplify and we'll
get:

x = x(x+2)

We'll divide
by x:

x+2=1

We'll subtract 2
both
sides:

x=-1

The
solution of the equation is x = -1.