Not sure how to do this. How do I integrate tan^2x?
D dagr8est Junior Member Joined Nov 2, 2004 Messages 128 Sep 9, 2006 #1 Not sure how to do this. How do I integrate tan^2x?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 9, 2006 #2 Derive the general reduction formula and you can use it for any power of tangent: \(\displaystyle \L\\\int{tan^{n}(x)}dx\\=\int{tan^{n-2}(x)tan^{2}(x)}dx\\=\int{tan^{n-2}(x)(sec^{2}(x)-1)}dx\\=\int{tan^{n-2}(x)sec^{2}(x)}dx-\int{tan^{n-2}(x)}dx\\=\frac{tan^{n-2}}{n-1}-\int{tan^{n-2}(x)}dx\)
Derive the general reduction formula and you can use it for any power of tangent: \(\displaystyle \L\\\int{tan^{n}(x)}dx\\=\int{tan^{n-2}(x)tan^{2}(x)}dx\\=\int{tan^{n-2}(x)(sec^{2}(x)-1)}dx\\=\int{tan^{n-2}(x)sec^{2}(x)}dx-\int{tan^{n-2}(x)}dx\\=\frac{tan^{n-2}}{n-1}-\int{tan^{n-2}(x)}dx\)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Sep 10, 2006 #3 Hello, dagr8est! \(\displaystyle \L\int\tan^2x\,dx\:=\:\int\left(\sec^2x\,-\,1)\,dx\) Can you finish it now?
Hello, dagr8est! \(\displaystyle \L\int\tan^2x\,dx\:=\:\int\left(\sec^2x\,-\,1)\,dx\) Can you finish it now?
