Evaluate the integral \[ \int tan 3x dx
V venialove Junior Member Joined Mar 30, 2008 Messages 53 Apr 20, 2008 #1 Evaluate the integral \[ \int tan 3x dx
skeeter Elite Member Joined Dec 15, 2005 Messages 3,218 Apr 20, 2008 #2 Re: Evaluate the integral ∫tanu du=−ln∣cosu∣+C\displaystyle \int \tan{u} \, du = -\ln|\cos{u}| + C∫tanudu=−ln∣cosu∣+C
Re: Evaluate the integral ∫tanu du=−ln∣cosu∣+C\displaystyle \int \tan{u} \, du = -\ln|\cos{u}| + C∫tanudu=−ln∣cosu∣+C
V venialove Junior Member Joined Mar 30, 2008 Messages 53 Apr 20, 2008 #3 Re: Evaluate the integral \(\displaystyle \int \tan{3x} \, dx This is how the problem is, but I don't know where you are getting the negative sign from? I got this answer: ln Icos 3xI +c\)
Re: Evaluate the integral \(\displaystyle \int \tan{3x} \, dx This is how the problem is, but I don't know where you are getting the negative sign from? I got this answer: ln Icos 3xI +c\)
skeeter Elite Member Joined Dec 15, 2005 Messages 3,218 Apr 20, 2008 #4 Re: Evaluate the integral ∫tanx dx=−∫−sinxcosx dx=−ln∣cosx∣+C\displaystyle \int \tan{x} \, dx = -\int \frac{-\sin{x}}{\cos{x}} \, dx = -\ln|\cos{x}| + C∫tanxdx=−∫cosx−sinxdx=−ln∣cosx∣+C
Re: Evaluate the integral ∫tanx dx=−∫−sinxcosx dx=−ln∣cosx∣+C\displaystyle \int \tan{x} \, dx = -\int \frac{-\sin{x}}{\cos{x}} \, dx = -\ln|\cos{x}| + C∫tanxdx=−∫cosx−sinxdx=−ln∣cosx∣+C
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Apr 20, 2008 #5 Re: Evaluate the integral ∫tan(3x)dx\displaystyle \int{tan(3x)}dx∫tan(3x)dx =∫sin(3x)cos(3x)dx\displaystyle =\int\frac{sin(3x)}{cos(3x)}dx=∫cos(3x)sin(3x)dx Let u=cos(3x), du=−3sin(x)dx, −du3=sin(x)dx\displaystyle u=cos(3x), \;\ du = -3sin(x)dx, \;\ \frac{-du}{3}=sin(x)dxu=cos(3x), du=−3sin(x)dx, 3−du=sin(x)dx =−ln(cos(3x))3+C\displaystyle = \frac{-ln(cos(3x))}{3}+C=3−ln(cos(3x))+C Since ln(cos(x))=−ln(1cos(x))\displaystyle ln(cos(x)) = -ln(\frac{1}{cos(x)})ln(cos(x))=−ln(cos(x)1) we can write it as ln(sec(3x))+C\displaystyle ln(sec(3x))+Cln(sec(3x))+C Technically, absolute values should be used, but.........
Re: Evaluate the integral ∫tan(3x)dx\displaystyle \int{tan(3x)}dx∫tan(3x)dx =∫sin(3x)cos(3x)dx\displaystyle =\int\frac{sin(3x)}{cos(3x)}dx=∫cos(3x)sin(3x)dx Let u=cos(3x), du=−3sin(x)dx, −du3=sin(x)dx\displaystyle u=cos(3x), \;\ du = -3sin(x)dx, \;\ \frac{-du}{3}=sin(x)dxu=cos(3x), du=−3sin(x)dx, 3−du=sin(x)dx =−ln(cos(3x))3+C\displaystyle = \frac{-ln(cos(3x))}{3}+C=3−ln(cos(3x))+C Since ln(cos(x))=−ln(1cos(x))\displaystyle ln(cos(x)) = -ln(\frac{1}{cos(x)})ln(cos(x))=−ln(cos(x)1) we can write it as ln(sec(3x))+C\displaystyle ln(sec(3x))+Cln(sec(3x))+C Technically, absolute values should be used, but.........
