# simple integration

• June 13th 2011, 08:40 PM
mjfuentes85
simple integration
ok I know how integrals work and I have been using integrals to find area under the curves, I know integration by U sub and also by parts. but for some reason maybe it just slipped my mind but how do I integrate:
integral tan (3x-1),x
I started off by doing a u sub as u = (3x-1) and where i got 1/3du = dx,
so my new equation became 1/3 int tan u du then i realized i don't know the integral of tan u. Can someone show me the steps using natural logs to the answer. I searched it on wolframaplpha but it gave me a log answer. Any help is appreciated.
-Marcus
• June 13th 2011, 08:50 PM
Also sprach Zarathustra
Re: simple integration
Quote:

Originally Posted by mjfuentes85
ok I know how integrals work and I have been using integrals to find area under the curves, I know integration by U sub and also by parts. but for some reason maybe it just slipped my mind but how do I integrate:
integral tan (3x-1),x
I started off by doing a u sub as u = (3x-1) and where i got 1/3du = dx,
so my new equation became 1/3 int tan u du then i realized i don't know the integral of tan u. Can someone show me the steps using natural logs to the answer. I searched it on wolframaplpha but it gave me a log answer. Any help is appreciated.
-Marcus

$\int \tan(3x-1)dx=-\int \tan(1-3x)dx \overset{1-3x=t}{=}\frac{1}{3}\int \tan(t)dt=-\frac{1}{3} \log|\cos(t)|+C=\log|\cos(1-3x)|+C$
• June 13th 2011, 08:52 PM
mjfuentes85
Re: simple integration
The answer I was trying to get was ln |sec x | + c, which is the integral of tanx dx , could you show me how to get this answer ??
• June 13th 2011, 09:15 PM
Also sprach Zarathustra
Re: simple integration
Quote:

Originally Posted by mjfuentes85
The answer I was trying to get was ln |sec x | + c, which is the integral of tanx dx , could you show me how to get this answer ??

$\log\cos(1-3x)=\log\frac{1}{\frac{1}{cos(1-3x)}}=(\log\sec(1-3x))^{-1}=-\log\sec(1-3x)$
• June 13th 2011, 09:19 PM
Prove It
Re: simple integration
Quote:

Originally Posted by Also sprach Zarathustra
$\log\cos(1-3x)=\log\frac{1}{\frac{1}{cos(1-3x)}}=(\log\sec(1-3x))^{-1}=-\log\sec(1-3x)$

You really need to get into the habit of writing modulus signs around the stuff inside logarithms you get through integrating...
• June 14th 2011, 04:07 AM
HallsofIvy
Re: simple integration
Quote:

Originally Posted by mjfuentes85
The answer I was trying to get was ln |sec x | + c, which is the integral of tanx dx , could you show me how to get this answer ??

$tan(x)= \frac{sin(x)}{cos(x)}$ so $\int tan(x)dx= \int\frac{sin(x)}{cos(x)}dx$. Let u= cos(x) so that du= -sin(x) and the integral becomes
$-\int \frac{du}{u}= -ln|u|+ C= -ln|cos(x)|+ C= ln|\frac{1}{cos(x)}|+ C= ln|sec(x)|+ C$