# Need Guidance...

• Sep 27th 2009, 05:57 PM
Juk3boxh3ro
Need Guidance...
Help with calculus problems needed? - Yahoo! Answers

I could use some advice im just looking for hints as to how to attack these problems... It's the beginning of Calc 2 and i didnt get the background i should have due to taking 1 in highschool with a not so great teacher.

Thank you
• Sep 27th 2009, 06:22 PM
Juk3boxh3ro
ok I figured out the x+1 over sqrt(16-x^2) one...

down to only looking for help on the other two
• Sep 27th 2009, 06:27 PM
Chris L T521
Quote:

Originally Posted by Juk3boxh3ro
ok I figured out the x+1 over sqrt(16-x^2) one...

down to only looking for help on the other two

$\displaystyle \int\frac{e^x\,dx}{e^{2x}-4}$

Make the subsitution $\displaystyle u=e^x$ to get $\displaystyle \int\frac{\,du}{u^2-4}$.

Can you take it from here?
• Sep 27th 2009, 06:52 PM
Juk3boxh3ro
is the answer 1/2 arc tan e^x / 2 + C??
• Sep 27th 2009, 07:02 PM
DeMath
Quote:

Originally Posted by Juk3boxh3ro
is the answer 1/2 arc tan e^x / 2 + C??

Not!!

$\displaystyle \int {\frac{{{e^x}}} {{{e^{2x}} - 4}}dx} = \left\{ \begin{gathered} {e^x} = u, \hfill \\ {e^x}dx = du \hfill \\ \end{gathered} \right\} = \int {\frac{{du}} {{{u^2} - 4}}} = \int {\frac{{du}} {{\left( {u - 2} \right)\left( {u + 2} \right)}}} .$

$\displaystyle \frac{1}{{\left( {u - 2} \right)\left( {u + 2} \right)}} = \frac{A}{{u - 2}} + \frac{B}{{u + 2}} \Leftrightarrow 1 = \left( {u + 2} \right)A + \left( {u - 2} \right)B.$

$\displaystyle {\text{If }}u = 2{\text{ then }}1 = 4A \Leftrightarrow A = \frac{1} {4}.$

$\displaystyle {\text{If }}u = - 2{\text{ then }}1 = - 4B \Leftrightarrow B = - \frac{1}{4}.$

$\displaystyle \int {\frac{{du}} {{{u^2} - 4}}} = \frac{1} {4}\int {\left( {\frac{1} {{u - 2}} - \frac{1} {{u + 2}}} \right)du} = \frac{1} {4}\left( {\ln \left| {u - 2} \right| - \ln \left| {u + 2} \right|} \right) + C =$

$\displaystyle = \frac{1} {4}\ln \left| {\frac{{u - 2}} {{u + 2}}} \right| + C = \frac{1} {4}\ln \left| {\frac{{{e^x} - 2}} {{{e^x} + 2}}} \right| + C.$
• Sep 27th 2009, 07:12 PM
Juk3boxh3ro
oh i see... sorry i used the wrong rule my notes are squeezed together and i lined them up wrong

thanks alot tho...

how about the algebraic form of tan (arc cos x/2)

i drew the right triangle x is on the adjacent side 2 is on the hypotenuse then you take the arc cos rule of u prime over square root of u^2 - 1

so...

the opposite side is 1/4 x^2 - 1??

idk sorry im so clueless...