1. ## [SOLVED] Trig Integral

$\int 3x sec (x) tan(x) dx$

Not sure what I need to use for my substitution.

2. $dv=\sec x\tan x dx \Rightarrow v=\sec x, u=3x\Rightarrow du=3dx$ and to calculate $\int \sec x dx$ rewrite as $\int \sec x \frac{\sec x + \tan x}{\sec x + \tan x}dx$

3. So I need to integration by parts?

4. Originally Posted by mollymcf2009
$\int 3x sec (x) tan(x) dx$

Not sure what I need to use for my substitution.
$\int 3x sec (x) tan(x) dx$

$u =secx == > arcsec(u) = x
$

$du = secxtanx$

$\int 3x sec (x) tan(x) dx$ = $3\int arcsec (u)du$

Integration by parts

$3\int arcsec (u)du$ = $\frac{3u}{arcsec (u)}$ - $3 \int \frac{1}{sqrt(u^2-1)}du$

Use trig substitution and i think you should get an answer.

5. Originally Posted by mollymcf2009
So I need to integration by parts?
Yes.

6. Here is what I am getting before I do any u substitutions or integration by parts:

$\int 3x \frac{sec^2xtanx +secxtan^2x}{secx + tanx} dx$

Is that what I need to do and then do my integraton by parts?

Edit 1: Wait,
Ok, so from my original problem, I need to so my integration by parts first, THEN when secx is in my integral from my integ. by parts, that is when I need to multiply top and bottom by secx + tanx
Right?

$3xsecx - 3 \ln|secx+tanx| + C$

Edit 2: GOT IT! Thanks y'all!

7. I did garbage before sorry about that i don't know what i was thinking.

$\int3xsecxtanxdx = 3xsecx - 3\intsecx$

secx = 1/cosx = cosx/ cos^2x = cosx/ (1 -sin^2x)

u = sinx
du =cosx

$\int3xsecxtanxdx = 3xsecx - 3\int1/(1-u^2)$

using partial fractions

1/(1-u^2) = $\frac{1/2}{1-u} + \frac{1/2}{1+u}$

$\int3xsecxtanxdx = 3xsecx - 3/2[ ln | 1+ sinx| +ln|1-sinx| ] +C$

Both answers are right although they appear different.