# Homework Help: Trig integral

1. Nov 8, 2006

### 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?

2. Nov 8, 2006

### 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!

Last edited: Nov 8, 2006
3. Nov 9, 2006

### dextercioby

Try the substitution $x=3\sinh t$.

Daniel.

4. Nov 9, 2006

### HallsofIvy

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

5. Nov 9, 2006

### 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.

$$2 \int sec^3(x) dx = sec(x)tan(x) + \int sec(x) dx$$
$$2 \int sec^3(x) dx = sec(x)tan(x) + ln|sec(x) + tan(x)| + K$$
$$\int sec^3(x) dx = {{sec(x)tan(x) + ln|sec(x) + tan(x)|}\over{2}} + C$$

How did I get
$$\int sec(x) dx = ln|sec(x) + tan(x)|$$?

Well, simplify this integral:

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

Last edited: Nov 9, 2006
6. Nov 9, 2006

### FunkyDwarf

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

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