Stuck in an Integration Problem: Help Needed!

[mbrmbrg]

I'm in the middle of solving \int\sqrt{x^2+9}dx and I got it into the form of 3\int\sec^3\theta d\theta, and I'm pulling a blank. Where does one begin? I could integrate by parts, setting u=sec(theta) and dv=sec^2(theta), but it's not getting me very far. Any suggestions?

 

[mbrmbrg]

bummer. i had that and decided it didn't look appealing...

OK, how's this?

3\int\sec^3\theta d\theta

=3(\sec\theta\tan\theta-\int\tan^2\theta\sec\theta d\theta)

=3(\sec\theta\tan\theta-\int\frac{sin^2\theta}{cos^3\theta}d\theta)

=3(\sec\theta\tan\theta-\int\frac{1+\cos^2\theta}{cos^3\theta}d\theta)

=3(\sec\theta\tan\theta-\int\frac{1+\cos^2\theta}{cos^3\theta}d\theta)

=3(\sec\theta\tan\theta-\int\sec^3\theta d\theta +\int\sec\theta d\theta)

Let I=\int\sec^3\theta d\theta

Then 3I=3(\sec\theta\tan\theta-\int\sec^3\theta d\theta +\int\sec\theta d\theta)

The 3's cancel, so I+I=(\sec\theta\tan\theta+\int\sec\theta d\theta)

so I=(\sec\theta\tan\theta+\int\sec\theta d\theta) and I can evaluate it from there (I hope).

Thank you very much!

Where did your post go, Courtigrad? I used it!

 

[dextercioby]

Try the substitution x=3\sinh t.

Daniel.

 

[HallsofIvy]

Here's how I would almost automatically do an integral like that. First convert secant to cosine
\int sec^3 x dx= \int \frac{1}{cos^3(x)}dx[/itex]<br /> which is an odd power of cosine. &quot;Take out&quot; a cos(x) to use with dx. That is, multiply both numerator and denominator by cos(x)<br /> \int \frac{cos (x)}{cos^4(x)}dx= \int \frac{cos(x)dx}{(1- sin^2(x))^2}<br /> Now let u= sin(x) so du= cos(x)dx<br /> \int \frac{du}{(1- u^2)^2}= \int \frac{du}{(1-u)^2(1+u)^2}<br /> and now use partial fractions.

 

[calcnd]

Here's how I integrated the \int sec^3(x)dx term.

\int sec^3(x)dx
=\int sec^2(x)sec(x)

u = sec(x)
du = sec(x)tan(x) dx

dv = sec(x)^2(x)dx
v = tan(x)\int sec^3(x)dx
= sec(x)tan(x) - \int tan(x)sec(x)tan(x) dx
= sec(x)tan(x) - \int tan^2(x)sec(x) dx
= sec(x)tan(x) - \int (sec^2(x) - 1)sec(x) dx
= sec(x)tan(x) - \int (sec^3(x) - sec(x) dx
= sec(x)tan(x) - \int (sec^3(x) + \int sec(x) dx

Now we have \int sec^3(x) on both sides, so consolidate them.

2 \int sec^3(x) dx = sec(x)tan(x) + \int sec(x) dx
2 \int sec^3(x) dx = sec(x)tan(x) + ln|sec(x) + tan(x)| + K
\int sec^3(x) dx = {{sec(x)tan(x) + ln|sec(x) + tan(x)|}\over{2}} + CHow did I get
\int sec(x) dx = ln|sec(x) + tan(x)|?

Well, simplify this integral:

\int sec(x) dx = \int sec(x)*{{sec(x)+tan(x)}\over{sec(x)+tan(x)}}

 

[FunkyDwarf]

i would have used the substitution x = 3 tan(t)