integral of cscx^(2/3)(cot^3)x
i know that cot^2x is csc^2(x)-1, but i just don't understand how to solve the cscx^(2/3), any help? i also know that its trig integrals/substitution...
let u = csc(x)
du = -csc(x) cot(x)
cot^2 = csc^2 - 1
and you have
csc^(2/3) (csc^2-1)cot(x)
= csc^(5/3)(csc*cot) - csc^(-1/3)(csc*cot)
= [-u^5/3 + u^(-1/3)] du
all downhill from there
To solve the integral ∫csc(x)^(2/3) * cot(x)^3 dx, we can make use of trigonometric substitutions. Let's start by considering the expression csc(x)^(2/3).
First, recall the trigonometric identity: csc^2(x) = cot^2(x) + 1. If we raise both sides to the power of 1/3, we get:
(csc^2(x))^(1/3) = (cot^2(x) + 1)^(1/3).
Now, we need to substitute a variable to simplify the expression. Let's use u = cot(x). This implies that du/dx = -csc^2(x), and dx = -du/csc^2(x).
Substituting these values into our integral expression, we have:
∫csc(x)^(2/3) * cot(x)^3 dx
= ∫csc(x)^(2/3) * (cot(x))^2 * cot(x) dx
= ∫(-csc(x)^(2/3) * u^2) * du
= - ∫u^2 * (csc(x)^(2/3)) * (-1/csc^2(x)) du
= ∫u^2 * csc^(-4/3)(x) du.
Now, we need to substitute back for x in terms of u. We know that u = cot(x), so x = arccot(u).
Differentiating both sides with respect to u gives:
dx/du = -(1/(1 + u^2)).
Now, we substitute these values into the integral expression:
∫u^2 * csc^(-4/3)(x) du
= ∫u^2 * csc^(-4/3)[arccot(u)] * dx/du du
= ∫u^2 * csc^(-4/3)[arccot(u)] * (-(1/(1 + u^2))) du.
This integral (u^2 * csc^(-4/3)[arccot(u)]) can be simplified by using another substitution. We can let w = arccot(u), so arccot(u) = w. Hence, the integral becomes:
-∫w^2 * csc^(-4/3)(w) dw.
At this point, you can proceed to solve this using numerical methods or approximation techniques. The final result will be in terms of the variable w.