# Recent content by n4rush0

1. ### 2 objects pulling on ice

Thanks, I wasn't sure if I could say the force on the man and force on the object both equal F.
2. ### 2 objects pulling on ice

Homework Statement Man with mass M pulling with constant horizontal force F along a rope attached to object with mass m. Both the man and the object are on a frictionless surface and separated by distance D. When the man and object meet, what is the velocity of the object? Velocity of the man...
3. ### Very Difficult Trig Sub?

Thank you. I finally get it now. I'll still have problems with the initial trig substitutions though since I'm not sure how to get tan(x/2) = t.
4. ### Very Difficult Trig Sub?

Thank you. I modified the integral to dt/t^2+t+1) Are you sure it's partial fractions?
5. ### Very Difficult Trig Sub?

I know how to change sinx to sin 2t/sqrt(1+4t^2) but I'm not sure how to simplify cos^2 (x/2) since it has the 1/2 in front of the x and I can't use the same trick that I used for sinx.
6. ### Very Difficult Trig Sub?

Okay so, given: integral dx/(2+sinx) tan(x/2) = t (1/2)sec^2 (x/2) dx = dt dx = 2cos^2 (x/2) dt integral 2cos^2 (x/2) dt / (2+sinx) Am I supposed to use x = arctan(2t)? If so, is it possible to simplify by drawing a triangle?
7. ### Very Difficult Trig Sub?

Where can I learn all these rules? I usually only see substitutions with x = asint, atant, or asect
8. ### Very Difficult Trig Sub?

Thanks, I'll try that. Is that something you just memorized or is there a certain rule that lets you know what to substitute?
9. ### Very Difficult Trig Sub?

Wow, that's so cool. Thanks for the link. I'll try to remember how to derive it.
10. ### Very Difficult Trig Sub?

Homework Statement Integral of 1/(2+sin(x)) dx Homework Equations The Attempt at a Solution I've been told that you can use trig subs, but I never had to learn that in high school and it hasn't appeared in any of my calculus coursework. As a side note. I've been wondering if it...
11. ### Integrating [1/(x^4 - x)]dx

Yes, that helps alot. Thanks, I solved it.
12. ### Integrating [1/(x^4 - x)]dx

I tried multiplying it out and got 1 = y^3 (A+B+C) + y^2 (B-C+D) + y(B-D) - A Even knowing that A = -1 and B = 1/3, I'm still not seeing the "nice" solution. Oh and, my original problem involved using y, but I wanted to use Mathmatica which solves dx integrals and I copied and pasted...
13. ### Integrating [1/(x^4 - x)]dx

So I set 1/[y(y-1)(y^2+y+1)] = A/y + B/(y-1) + (Cy+D)/(y^2+y+1) 1 = (y-1)(y^2+y+1)A + y(y^2+y+1)B + y(y-1)(Cy+D) y = 0, A = -1 y = 1, B = 1/3 Not sure what to do for C and D
14. ### Integrating [1/(x^4 - x)]dx

Homework Statement [1/(x^4 - x)]dx Homework Equations The Attempt at a Solution I factored the denominator to x(x-1)(x^2 + x +1) and I'm not sure if I can use partial fractions.