G
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,216
atse1900 said:
like "they" get everything else in math that you fail to understand ... friggin' magic.
:wink:
not really ...
some person once noticed that secx(secx + tanx) = sec<sup>2</sup> + secxtanx is the derivative of secx + tanx ... hence the form f'(x)/f(x) which, when integrated, is ln|f(x)| ... ln|secx + tanx| in this case.
ChaoticLlama
Junior Member
- Joined
- Dec 11, 2004
- Messages
- 199
you're right, that multiplying by (secx + tanx)/(secx + tanx) looks pretty arbitrary, but it's just the common short-cut that is shown.
Another way to do this integral yourself is to rewrite it like this:
secx
= 1/cosx
= cosx/cos²x
= cosx/(1-sin²x)
On that last line, let something like z=sinx, and then use partial fractions.
Another way to do this integral yourself is to rewrite it like this:
secx
= 1/cosx
= cosx/cos²x
= cosx/(1-sin²x)
On that last line, let something like z=sinx, and then use partial fractions.