How do I go about integrating sin[x^(1/3)]dx?? I tried doing integration by parts and I'm going in circles.
let x^(1/3) = w
w^3 = x
dx = 3w^2 dw
The integral becomes
3 *INTEGRAL w^2 sin w dw
The integral of that is in my table of integrals and it looks like you need to apply integration by parts twice to get it.
I GOT IT! Thank you so so much for guiding me in the right direction!!
I'm sure glad you had the satisfaction of solving it; I was just too lazy to do the whole thing.
To integrate the function sin[x^(1/3)]dx, you can try a different approach by making a substitution. Let's start from the beginning:
1. First, let's make a substitution to simplify the integral. Let u = x^(1/3). This implies du/dx = (1/3)x^(-2/3). Rearranging the equation, we get dx = 3u^2du.
2. Now substitute these values into the integral. We have:
∫sin[x^(1/3)]dx = ∫sin(u) * 3u^2 du.
3. Next, we can use integration by parts to integrate this new function ∫sin(u) * 3u^2 du. The formula for integration by parts is:
∫u * dv = u * v - ∫v * du,
where u and v are functions of u, and du and dv are their derivatives with respect to u.
4. Let's assign u = 3u^2 and dv = sin(u) du. Taking derivatives, we find du = 6u du and v = -cos(u).
5. Using the integration by parts formula, we can evaluate the integral:
∫sin(u) * 3u^2 du = (3u^2)(-cos(u)) - ∫(-cos(u))(6u du).
6. The next step is to simplify and integrate the remaining integral:
= -3u^2cos(u) + 6∫u cos(u) du.
7. At this point, we are left with another integral ∫u cos(u) du, which requires further evaluation.
To solve this integral, you can again use integration by parts by assigning u = u and dv = cos(u) du. By taking derivatives, you get du = du and v = sin(u).
8. Applying the integration by parts formula once again, we have:
∫u cos(u) du = (u)(sin(u)) - ∫sin(u) du.
9. The new integral reduces to ∫sin(u) du, which we know is equal to -cos(u). Hence:
∫u cos(u) du = (u)(sin(u)) - ∫sin(u) du
= (u)(sin(u)) + cos(u) + C,
where C is the constant of integration.
10. Plugging this result back into our previous integral, we have:
-3u^2cos(u) + 6∫u cos(u) du = -3u^2cos(u) + 6[(u)(sin(u)) + cos(u)] + C.
11. Finally, substitute back u = x^(1/3) to get the final answer:
∫sin[x^(1/3)]dx = -3(x^(1/3))^2cos(x^(1/3)) + 6[x^(1/3)sin(x^(1/3)) + cos(x^(1/3))] + C.
Please note that C represents the constant of integration and should be included at the end of the answer.