Can someone please show me the exact steps to finding the F(x) of:

y = {sec x tan x dx

Can someone please show me the exact steps to finding the F(x) of:

y = {sec x tan x dx

 

Do you know the trig identities?

Okay, there aren't many steps to show for this one. You just have to realize that the derivative of Sec(x) is Sec(x)Tan(x), so your answer is just Sec(x)+C

if you sexy(sec) you have a sexy tan (sec tan)

 

if you have a tan your sexy squared. (tan --> sec sec)

 

 

: )

J33bu5, so it's almost memorization!! I must've been thinking too hard...

You can also easily show it using the trig identities like revprez pointed out. It works out fairly simple (you said you wanted to do it by hand so I assumed you meant with steps involved)

 

Ssin[x]/(cos[x])^2 dx u=cos[x] du=-sin[x]

 

-S (1/u^2) du

=-(-1/u) + c

=1/cos[x]+c= sec[x] + c

 

Simple as that