Question

integral of sec^2((pi/16)x)-1dx

Answer 1

Hi, welcome to OpenStudy.\[\Large\rm \int\limits \sec^2\left(\frac{\pi}{16}x\right)-1~~dx\]So we need to integrate this?

Answer 2

Do you remember this integral? \(\Large\rm \int sec^2x~dx\)
Or maybe I should ask... do you remember this derivative? \(\Large\rm \dfrac{d}{dx}\tan x=\sec^2x\)

Answer 3

yes i remember that but doesn't the -1 effect it?

Answer 4

\[\int sec^2(something)-1 dx=\int sec^2(something)dx -\int 1dx\]