the integration of : sec[x]^3 i know the answer just cant figure out how to get there pleaze help
N naz786 New member Joined Dec 15, 2005 Messages 3 Feb 20, 2006 #1 the integration of : sec[x]^3 i know the answer just cant figure out how to get there pleaze help
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Feb 20, 2006 #2 OK, here goes: Let u=sec(x)\displaystyle u=sec(x)u=sec(x),dv=sec2(x)dx\displaystyle dv=sec^{2}(x)dxdv=sec2(x)dx, \(\displaystyle du=sec(x)tan (x)dx\), v=tan(x)\displaystyle v=tan(x)v=tan(x). \(\displaystyle \L\int{sec^{3}(x)}dx=sec(x)tan(x)-\int{sec(x)tan^{2}(x)dx}\) =\(\displaystyle \L\sec(x)tan(x)-\int{sec(x)(sec^{2}(x)-1)dx\) =\(\displaystyle \L\sec(x)tan(x)-\int{sec^{3}(x)dx}+\int{sec(x)dx\), \(\displaystyle \L\2\int{sec^{3}(x)dx=sec(x)tan(x)+\int{sec(x)dx\), Therefore, \(\displaystyle \L\int{sec^{3}(x)}dx=\frac{1}{2}sec(x)tan(x)+\frac{1}{2}\int{sec(x)dx\) =\(\displaystyle \L\int{sec^{3}(x)dx=\frac{1}{2}sec(x)tan(x)+\frac{1}{2}ln(|sec(x)+tan(x)|)+C\)
OK, here goes: Let u=sec(x)\displaystyle u=sec(x)u=sec(x),dv=sec2(x)dx\displaystyle dv=sec^{2}(x)dxdv=sec2(x)dx, \(\displaystyle du=sec(x)tan (x)dx\), v=tan(x)\displaystyle v=tan(x)v=tan(x). \(\displaystyle \L\int{sec^{3}(x)}dx=sec(x)tan(x)-\int{sec(x)tan^{2}(x)dx}\) =\(\displaystyle \L\sec(x)tan(x)-\int{sec(x)(sec^{2}(x)-1)dx\) =\(\displaystyle \L\sec(x)tan(x)-\int{sec^{3}(x)dx}+\int{sec(x)dx\), \(\displaystyle \L\2\int{sec^{3}(x)dx=sec(x)tan(x)+\int{sec(x)dx\), Therefore, \(\displaystyle \L\int{sec^{3}(x)}dx=\frac{1}{2}sec(x)tan(x)+\frac{1}{2}\int{sec(x)dx\) =\(\displaystyle \L\int{sec^{3}(x)dx=\frac{1}{2}sec(x)tan(x)+\frac{1}{2}ln(|sec(x)+tan(x)|)+C\)
