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western50 · Sep 8, 2010, 11:31 PM

how to integrate this

(sec x)(tan x)^2

galactus · Sep 9, 2010, 01:12 AM

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

Rewrite:

$\int (\underbrace{sec^{2}(x)-1}_{\text{tan^2(x)}})sec(x)dx$

$\int sec^{3}(x)dx-\int sec(x)dx$

Now, integrate each one. The first one can be done by just using the reduction formula. The second is just $ln|sec(x)+tan(x)|+C$ , and can be derived by letting $u=sec(x)+tan(x), \;\ du=(sec^{2}(x)+sec(x)tan(x))dx$

Reduction formula:

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

The reduction formula can be derived by using integration by parts.