Question

ln(secx+tanx)+C = integral (secx)dx
I don't really understand why...

Answer 1

\[\ln (\sec x+ \tan x) +C = \int\limits \sec x dx\]

Answer 2

would it help to write 1/cos x in for sec x ?

Answer 3

That's one way, and change the cos(x) in the denominator to \[\sin{\frac{\pi}{2} + x}\], etc. but another way is multiply sec(x) by \[\frac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)}\]

Answer 4

Do exactly what he said multiply by (secx+tanx)/(secx+tanx) and then do u substitution. make the denominator your u. When you take du you will find it is the numerator.

Answer 5

so i'm assuming you would do this for the integral side?

Answer 6

ohhh ok thanks :)

Answer 7

an /alternative/ approach would be to simply differentiate the left-hand-side and show that it equals sec(x)

Answer 8

@jennychan12 Yes!

Answer 9

lol i would do that, but we're doing integrals right now :(

Answer 10

fair enough - then u substitution it is as suggested by the others :)

Answer 11

yes, i got the answer. well. i verified the answer. thanks :)

Answer 12

Yw.