# integration problems

• Mar 17th 2009, 07:18 AM
scottie.mcdonald
integration problems
i'm having trouble (as always), starting these problems. hints on how to start would be nice.

evaluate the integrals

a). $\displaystyle \int sin\sqrt(at)dt$

i think i have to do this by integration by parts, so i have:

$\displaystyle f(x) = sin \sqrt(at), f\prime(x) = -cos \sqrt(at), g(x)=t, g\prime(x)=dt$.

though i think this is wrong.

b). $\displaystyle \int sin4xcos3xdx$.

again, i think it's integration by parts, but the 4 and 3 are messing me up

c). $\displaystyle \int \sqrt(1+e^x)dx$

i have $\displaystyle f(x)=1+e^x, f\prime(x)=xe^x, g(x)=x, g\prime(x)=dx$

the square root is messing me up, so again, i don't think i selected the parts correctly.

any help on how to get started would be nice.

thank you,

Scott
• Mar 17th 2009, 08:26 AM
Soroban
Hello, scottie.mcdonald!

The third one requires a "simple" substitution . . .

Quote:

$\displaystyle c)\;\;\int \sqrt{1+e^x}\,dx$

Let: .$\displaystyle u \:=\:\sqrt{1+e^x} \quad\Rightarrow\quad u^2 \:=\:1 + e^x \quad\Rightarrow\quad e^x \:=\:u^2-1 \quad\Rightarrow\quad x \:=\:\ln(u^2-1)$

. . Then: .$\displaystyle dx \:=\:\frac{2u}{u^2-1}\,du$

Substitute: .$\displaystyle \int u \left(\frac{2u}{u^2-1}\,du\right) \;=\;2\int\frac{u^2}{u^2-1}\,du$

. . $\displaystyle =\;2\int\left(1 + \frac{1}{u^2-1}\right)\,du \;=\;2\left[u + \frac{1}{2}\ln\left(\frac{u-1}{u+1}\right)\right] \;+\; C \;=\;2u \;+ \;\ln\left(\frac{u-1}{u+1}\right) \;+\; C$

Back-substitute: .$\displaystyle 2\sqrt{1+e^x} + \ln\left(\frac{\sqrt{1+e^x} - 1}{\sqrt{1+e^x} + 1}\right) + C$ .[1]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We could stop here, but we can simplify the answer further . . .

Multiply the fracton by $\displaystyle \frac{\sqrt{1+e^x} -1}{\sqrt{1+e^x}-1}$

. . $\displaystyle \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x} + 1} \cdot \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x} - 1} \;=\;\frac{(\sqrt{1+e^x}-1)^2}{(1+e^x) - 1} \;=\;\frac{(\sqrt{1+e^x} -1)^2}{e^x}$

Then [1] becomes: .$\displaystyle 2\sqrt{1+e^x} + \ln\left[\frac{(\sqrt{1+e^x} - 1)^2}{e^x}\right] + C$

. . . . . . . . . . . . $\displaystyle = \;2\sqrt{1+e^x} + \ln(\sqrt{1+e^x} - 1)^2 - \ln(e^x) + C$

. . . . . . . . . . . . $\displaystyle = \;2\sqrt{1+e^x} + 2\ln(\sqrt{1+e^x}-1) - x + C$

Pull something like this on any of your exams,
. . and I'll guarentee that you'll surprise/impress/terrify your teacher.

• Mar 17th 2009, 12:45 PM
scottie.mcdonald
ha ha yeah, my teacher would look at me with wonder if i thought about rationalizing at the end (though now i'm always going to look for that).

Thanks for the help =)