For my second problem, my upper limit was square root of e+1.
And I know this may seem like a dumb question, but when you move 1/2 behind the integration sign, how do you know when to do that? I realize it is there to cancel out the 2 that is placed in front of the function, but is it due to the aftermath of the substitution rule?
There are two slightly different ways to think about that.
Your integral is and you want to use the substitution . Then so
Now either
1) Recognize that you have in the numerator but need so multiply and divide by 2:
or
2) Divide both sides of by 2 to get so that
.
Note that only constants can be moved out of the integral like that. If the "x" had not been in the numerator to begin with, we could not have used that substitution.
Thank you Mr. Fantastic and Prove it. I think I understand it now. I have done the problem over again, and hopefully I have done it correctly this time!
EDIT: TYPO! The answer is 1/2, I realized I made a small error on the bottom when I had 1/2(lne). I made it equal to 1, when in reality it is 1/2*1=1/2.
For the 3rd problem, you have to know the anti-derivative of the trig function cos(x).
The anti-derivative for cos(x) is sin(x), but in this case we have a cos(2x) which makes finding the anti-derivative a not any harder to be honest. The anti-derivative for cos(2x) is sin(2x)/2
basically your just taking the coefficient of x (which is 2) and dividing it by the anti-derivative and this works in pretty much any trig function.
so in any case say finding the anti-derivative of sin(Cx) would equal -cos(Cx)/C where C is represents the coefficient.
Thanks! I redid that problem and I think I get it now!
EDIT: Just one small complication though. I checked the answers in the back of the book, and it has the same answer as me only positive, while mine is negative. I'm not sure why though.
EDIT [again]: Nevermind, I just realized I calculated cos's antiderivative incorrectly; I put -sin rather than positive sin.