100fg per problem if you can show work, ive already worked B out but am getting a different answer as the instructors solution. Just need someone to verify

Please integrate B by parts uv-int(vdu) etc...

As far as A its been too long and cant remember where to begin i know its simple....




http://i.imgur.com/2SwQh3c.jpg?1?1909

This post was edited by kookymunsta309 on Dec 5 2013 06:37pm

doing it now, dont send fg to anyone else plz
tried to send you but your inbox is full...posting it below

had it done before saber

This post was edited by JDota72 on Dec 5 2013 07:06pm

A is actually very simple
It's a simple u substitution

derivative of cos(x)= -sin(x)
derivative of e = 0, constant
derivative of ln(secx)= 1/(secx) = tan(x)
so it cancels the top

part A

u = cos(x) + ln(sec(x)) + e
du/dx = tan(x) - sin(x)
1/(tan(x) - sin(x)) du = dx

using that:
integral (tan(x) - sin(x)/u) * 1/(tan(x) - sin(x)) du
integral (1/u)*du
integral u^(-1)* du

=ln(|u|) + C
=ln(|cos(x) + ln(sec(x)) + e|) + C

This post was edited by Rocinante on Dec 5 2013 07:26pm

Part b, maybe a typo?

The tan(x^2) makes it pretty much not solvable by integration by parts, or rather difficult requiring multiple integration by parts

maybe its x*[tan(x)]^2

that could be done by parts

part B
assuming x*[tan(x)]^2

integral x*tan^2(x)*dx

integral x*(sec^2(x)-1)*dx

integral [x*sec^2(x) - x]*dx

integral x*sec^2(x)*dx - integral x*dx

f=x
df=dx
dg=sec^2(x)*dx
g=tan(x)

integral x*sec^2(x)*dx - integral x*dx = x*tan(x) - integral tan(x)*dx - integral x*dx
right hand side is solvable, so lets do that:
=x*tan(x) + ln(cos(x)) - (x^2)/2 + C

This post was edited by Rocinante on Dec 5 2013 07:48pm

Thanks, Yep same answer as mine just wanted to verify he had made a mistake on our sheet

This post was edited by kookymunsta309 on Dec 5 2013 07:52pm