[((cotx)^(1/3))((csc^4(x))]

i don't understand how to integrate the first part because it should be either even or odd, but this is neither.. help!

let u = cot(x)

du = -csc^2(x) dx

csc^4 = csc^2 * (1+cot^2)
and we have

-u^(1/3)(1+u^2) du

That should pose no problem.

To integrate the expression ((cot(x))^(1/3))(csc^4(x)), we can start by rewriting the trigonometric functions in terms of sine and cosine.

Using the Pythagorean identity, csc(x) = 1/sin(x) and cot(x) = 1/tan(x) = cos(x)/sin(x), we can rewrite the expression as:

((cos(x)/sin(x))^(1/3))(1/sin^4(x))

Next, consider the exponent on the cot term, which is 1/3. To integrate this, let's make a substitution u = sin(x).

Differentiate u with respect to x to get du = cos(x) dx, or dx = du / cos(x).

Now let's rewrite the expression with this substitution:

((cos(x)/sin(x))^(1/3))(1/sin^4(x)) = ((1/u)^(1/3))(1/u^4) = u^(-1/3) u^(-4) = u^(-13/3)

Now we need to integrate u^(-13/3) with respect to u.

Integrating, we get:

∫ u^(-13/3) du

To integrate this term, we can use the power rule for integration:

∫ x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration.

Applying this rule to our integral, we have:

∫ u^(-13/3) du = (u^(-13/3 + 1))/(-13/3 + 1) + C

Simplifying, we get:

∫ u^(-13/3) du = (u^(-10/3))/(-10/3) + C

Finally, substituting u back in terms of x, we have:

∫ ((cot(x))^(1/3))(csc^4(x)) dx = (sin^(-10/3)(x))/(-10/3) + C

Therefore, the integration of ((cot(x))^(1/3))(csc^4(x)) with respect to x is (sin^(-10/3)(x))/(-10/3) + C.