integrate (secx)^4

**Bryn** · June 15th 2008, 06:05 AM

integrate (secx)^4

Please help,

Thanks
Bryn

**kalagota** · June 15th 2008, 06:11 AM

Originally Posted by Bryn

integrate (secx)^4

Please help,

Thanks
Bryn

$\int \sec^4x \, dx = \int (\sec^2x)\sec^2x \, dx = \int (1 + \tan^2x)\sec^2x \, dx$

set $u = \tan x$ ...

**galactus** · June 15th 2008, 06:11 AM

If you want the easy way out, use the general formula:

$\int{sec^{n}(u)}du=\frac{1}{n-1}sec^{n-2}(u)tan(u)+\frac{n-2}{n-1}\int{sec^{n-2}(u)}du$

or proceed:

$\int{sec^{4}(x)}dx=\int(1+tan^{2}(x))sec^{2}(x)dx$

$=\int(sec^{2}(x)+tan^{2}(x)sec^{2}(x))dx$

$=\int{tan^{2}(x)}dx+\int{tan^{2}(x)sec^{2}(x)}dx$

The second half let $u=tan(x), \;\ du=sec^{2}(x)dx$

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