Integration: Evaluate the Integral?

[KAISER91]

Homework Statement

Evaluate;

∫ x sec^2 x dx

Homework Equations

The Attempt at a Solution

So far this is what I have;

u = x
du/dx = 1
du = 1 dx


dv = sec^2 x
v = tan x


Therefore;

= x tan x - ∫ tan x dx
= x tan x - ln (sec x) + c




I stopped there thinking that was the answer?


The answer given is

x tan x + ln (cos x) + c



Thanks

 

[HallsofIvy]

Where did you get that
\int tan(x)dx= ln(sec(x))+ C?

\int tan(x)dx= \int \frac{sin(x)}{cos(x)}dx
Let u= cos(x) so du= sin(x)
\int tan(x)dx= \int \frac{du}{u}= ln(u)+ C= ln(cos(x))+ C

 

[cepheid]

Yeah, the answer given is correct. You must have just integrated tanx incorrectly. Hint: this integral is done easily by substitution

 

[George Jones]

Therefore;

= x tan x - ∫ tan x dx
= x tan x - ln (sec x) + c




I stopped there thinking that was the answer?


The answer given is

x tan x + ln (cos x) + c

What is another way of expressing ln (sec x)?

 

[KAISER91]

Where did you get that
\int tan(x)dx= ln(sec(x))+ C?

\int tan(x)dx= \int \frac{sin(x)}{cos(x)}dx
Let u= cos(x) so du= sin(x)
\int tan(x)dx= \int \frac{du}{u}= ln(u)+ C= ln(cos(x))+ C

OH.Thanks. I appreciate the help.

Note, you forgot the negative sign on sin x for the derivative of cos x.Thanks.

 

[KAISER91]

What is another way of expressing ln (sec x)?

sin x / cos xLOLOL. I totally forgot about it.

 

[George Jones]

sin x / cos x

I'm not sure what you mean.

sec x = ?

ln(sec x) = ?