# can someone check this calculus problem for me?

Discussion in 'Community Discussion' started by macman2790, Oct 25, 2007.

1. ### macman2790 macrumors 6502a

Joined:
Sep 4, 2006
Location:
Texas
#1
This was on a test and this is one of the problems i got "wrong". I think that my calculation is right, because of the formula she gave us to do this one.
the formula is integral(sec^n(x)dx )= 1/(n-1)sec^(n-1)xtanx + (n-2)/(n-1)integral(sec^(n-2)xdx)
integrate sec^5(x)
I did this problem the correct way:First i did this formula twice since It gave me the integral sec^3 x. then i integrated sec(x) which is given at the end of the second cycle and it gave me the answer. I've tried this problem many times.
Here is my answer and it makes no sense to me why it is wrong. a 1/2 coefficient was counted wrong but i dont get why.
1/4sec^4(x)tan(x)+3/8sec^2(x)tan(x)+1/2ln|sec(x)+tan(x)| +c.
the last coefficent(1/2ln|sec(x) + tan(x) | was counted wrong andit makes no sense why because the formula ends up putting n-1/n-2 before the last integral. and n is equal to three in that part. can someone tell me if i'm right or the professor is right? thanks.

2. ### siurpeeman macrumors 603

Joined:
Dec 2, 2006
Location:
the OC
#2
sorry, the professor's got you on this one. you forgot to distribute the 3/4 to the integral of (sec x)^3, so that last 1/2 should be another 3/8.

3. ### Rodimus Prime macrumors G4

Joined:
Oct 9, 2006
#3
let me guess you are in the area doing trig subsition.

word of advice about cal. Get really good at your trig proofs. They will save you.

Also remember cal is 90% algebra 10% cal

Joined:
Sep 4, 2006
Location:
Texas