Recent content by Ianfinity

The question is how do I prove the reduction formula here. I used u=(sec x)^n and dv=dx, so du=sec(x)tan(x)dx and v=x Where I'm getting most confused at is the part where it says to add (n-2)/(n-1)int(sec^(n-2))xdx I don't see why (n-2)/(n-1) is multiplied by the integral there... I...

Use integration by parts to prove the reduction formula: int(sec^n)x dx = (tan(x)*sec^(n-2)*x)/(n-1) + [(n-2)/(n-1)]int(sec^(n-2)*x dx n /= 1 (n does not equal 1) I used "int" in place of the integral sign. This was a problem on the corresponding test from the cal A class I am from...

Yes, I am sure. It IS possible, however, that the problem has no solution, if that's what you're getting at.

Evaluate cos(arccos((32pi)/3)) From my understanding, I need to subtract (32pi)/3 by pi until the answer falls within the domain of [-1, 1]. Is this the correct way to solve this problem?

g(x) = 3 if x < -4 g(x) = 7+x if |x| <or= 4 g(x) = x^4 if x > 4 I know there is a jump discontinuity at x=4. How would I state that in interval notation? Is that even possible or is it good enough to say g(x) is discontinuous at x=4? Basically what I've found is that if x=4 then g(x)=11...