Whatis the integration of ∫▒〖 ( dx)/(〖sin〗^4 x+〖sin〗^2 x 〖cos〗^2 x+cos^4x )〗
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!
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
To find the integration of the given expression, we can use a method called partial fraction decomposition. This involves breaking down the fraction into simpler fractions that are easier to integrate.
Let's start with the given expression:
∫[(dx)/(sin^4(x) + sin^2(x)cos^2(x) + cos^4(x))]
To simplify the denominator, we can factorize it as follows:
sin^4(x) + sin^2(x)cos^2(x) + cos^4(x) = (sin^2(x) + cos^2(x))^2 - sin^2(x)cos^2(x)
Using the identity sin^2(x) + cos^2(x) = 1, we can simplify further:
(sin^2(x) + cos^2(x))^2 - sin^2(x)cos^2(x) = 1 - sin^2(x)cos^2(x)
Now, we can substitute this back into the original expression:
∫[(dx)/(1 - sin^2(x)cos^2(x))]
To perform the partial fraction decomposition, let's factorize the denominator:
1 - sin^2(x)cos^2(x) = (1 - sin(x)cos(x))(1 + sin(x)cos(x))
Now, we can rewrite the integral using partial fraction decomposition:
∫[(dx)/((1 - sin(x)cos(x))(1 + sin(x)cos(x)))]
To proceed further, we'll need to apply the method of partial fraction decomposition. This involves expressing the expression as a sum of two simpler fractions:
∫[(dx)/(1 - sin(x)cos(x))(1 + sin(x)cos(x))] = ∫[(A/(1 - sin(x)cos(x))) + (B/(1 + sin(x)cos(x)))] dx
First, let's find the values of A and B. We'll need a common denominator:
A(1 + sin(x)cos(x)) + B(1 - sin(x)cos(x)) = 1
Expanding the left side:
A + A*sin(x)cos(x) + B - B*sin(x)cos(x) = 1
Grouping like terms:
(A + B) + (A - B)sin(x)cos(x) = 1
Comparing coefficients of sin(x)cos(x):
A - B = 0 (1)
A + B = 1 (2)
From equation (1), we can see that A = B. Substituting this into equation (2), we get A = B = 1/2.
Now, we can rewrite the integral again using the values of A and B:
∫[(1/2)/(1 - sin(x)cos(x))] + [(1/2)/(1 + sin(x)cos(x))] dx
Splitting the integral into two:
(1/2)∫[(1/(1 - sin(x)cos(x))) + (1/(1 + sin(x)cos(x)))] dx
Now, we can integrate each term separately:
(1/2)∫(dx/(1 - sin(x)cos(x))) + (1/2)∫(dx/(1 + sin(x)cos(x)))
The integrals in both terms have the form dx/(1 - u^2) and dx/(1 + u^2), where u = sin(x)cos(x) for respective terms.
The integrals of the form dx/(1 - u^2) and dx/(1 + u^2) can be solved by using trigonometric substitutions, specifically using the substitution u = sin(theta) and u = tan(theta) respectively.
I hope this explanation helps!