integrate (secx)^4
Please help,
Thanks
Bryn
If you want the easy way out, use the general formula:
$\displaystyle \int{sec^{n}(u)}du=\frac{1}{n-1}sec^{n-2}(u)tan(u)+\frac{n-2}{n-1}\int{sec^{n-2}(u)}du$
or proceed:
$\displaystyle \int{sec^{4}(x)}dx=\int(1+tan^{2}(x))sec^{2}(x)dx$
$\displaystyle =\int(sec^{2}(x)+tan^{2}(x)sec^{2}(x))dx$
$\displaystyle =\int{tan^{2}(x)}dx+\int{tan^{2}(x)sec^{2}(x)}dx$
The second half let $\displaystyle u=tan(x), \;\ du=sec^{2}(x)dx$