Whatis the integration of ∫▒〖 ( dx)/(〖sin〗^4 x+〖sin〗^2 x 〖cos〗^2 x+cos^4⁡x )〗

Answer 1

wow - some font magic there. I'll go with

∫dx/(sin^4 x +sin^2 x cos^2 x + cos^4⁡ x)
Note that 1 = (sin^2x + cos^2x)^2 = sin^4x+2sin^2x*cos^2x+cos^4x

So the integral can be written as

∫dx/(1-sin^2 x cos^2 x)
= ∫dx/(1-sin^2x(1-sin^2x))
= ∫dx/(1-sin^2x+sin^4x)
= ∫dx/(cos^2x+sin^4x)
= ∫ sec^4x dx / (sec^2x+tan^4x)
= ∫ sec^2x(1+tan^2x) dx / (1+tan^2x+tan^4x)

Now let
t = tanx
dt = sec^2x dx

and the integral is now

∫ (1+t^2)/(1+t^2+t^4) dt
= ∫(1 + 1/t^2)/(1 + 1/t^2 + t^2) dt
= ∫(1+1/t^2)/((t-1/t)^2+3) dt

Now let
u = t - 1/t
du = (1+1/t^2) dt

and the integral is now

∫du/(u^2+(√3)^2)

Now recall that one of the standard formulas for integrals is

∫ du/(u^2+a^2) = 1/a arctan(u/a)

and we have

∫du/(u^2+(√3)^2) = 1/√3 arctan(u/√3)
= 1/√3 arctan(1/√3 (tanx-cotx))
= 1/√3 arctan(√3/2 tan(2x)) + C

whew!

Answer 2

Hmmm. The intermediate t-stuff might be avoided if we proceed thusly:

(tanx-cotx)^2 = tan^2x - 2 + cot^2x
= sin^2x/cos^2x - 2 + cos^2x/sin^2x
= (sin^4x - 2sin^2x*cos^2x + cos^4x)/(sin^2x cos^2x)
= (sin^4x+sin^2x*cos^2x+cos^4x - 3sin^2x*cos^2x)/(sin^2x cos^2x)

so

sin^4x+sin^2x*cos^2x+cos^4x = sin^2x*cos^2x (tanx-cotx)^2 + 3)

Now, if we let
u = tanx-cotx
du = (sec^2x+csc^2x) dx
= 1/(sin^2x cos^2x)
= sec^2x csc^2x dx

Now we have our original integrand as

sec^2x csc^2x dx / ((tanx-cotx)^2 + 3)
= du/(u^2+3)

as above