# Thread: Evaluating integrals with antiderivatives

1. ## Evaluating integrals with antiderivatives

Hi, I'm trying to study for a quiz about evaluating integrals using antiderivatives. I understand that F(x)+C = int[f(x) dx] but some problems I'm not sure where to go as to solving it.

1) int[2secxtanx]
I thought the answer would be 2ln(cosx)+C
But my book says the answer of the antiderivative would be tan5x+C

2) int[(tanx)^2]
I thought the answer would be (1/3)[(tanx)^3]+C
But the book says it's tanx-x+C

I'm doing something wrong here but I don't know what... It would be great if someone could explain these for me and give any tips of an efficient method of finding antiderivatives. (And is it possible to use the substitution method in these types of integrals?)

2. 1) think of the problem as this:
$sec(x) = 1/cos(x)$ and
$tan(x) = sin(x)/cos(x)$

so the problem becomes
$
\int sin(x)/cos(x)^2
$

set u = cos(x) and du = -sin(x)dx

integral is then:

$
\int 1/u^2$

the end result should be $2/cos(x) + c$ and not tan(5x)

2) tan(x)^2 is really $sin(x)^2/cos(x)^2$

first you need to use the trig identity $sin(x)^2 + cos(x)^2 = 1$ and replace the sin(x)^2

then it becomes $(1-cos(x)^2)/cos(x)^2$

separate each $1/cos(x)^2 - 1$

which is then $\int sec(x)^2 - 1$

$\int sec(x)^2 = tan(x)$ and $\int -1 = -x$

end result being $tan(x) - x + c$

3. Originally Posted by deltahunter
1) think of the problem as this:
$sec(x) = 1/cos(x)$ and
$tan(x) = sin(x)/cos(x)$

so the problem becomes
$
\int sin(x)/cos(x)^2
$

set u = cos(x) and du = -sin(x)dx

integral is then:

$
\int 1/u^2$

the end result should be $2/cos(x) + c$ and not tan(5x)
To make it a little easier, you could note that $\frac{\,d}{\,dx}\left(\sec x\right)=\sec x\tan x\implies \int\sec x\tan x\,dx=\sec x+C$

4. Yea, I just wanted to go through it and show that it's not tan(5x)