The average value of the function $y=\tan(2x)$ over the interval $[0,\frac{\pi}{8}]$

Question

I was given the following question in a technology free exam. How would one go about solving this without the use of a calculator?

_{NB. I am currently in my last year of high school so please don't suggest crazy equations that contain so many greek symbols it could be a greek dictionary - Unless of course that is the only way to complete the question.}

The average value of the function $y=\tan(2x)$ over the interval $[0,\frac{\pi}{8}]$ is
A. $\frac{2}{\pi}\log_e({2})$
B. $\frac{\pi}{4}$
C. $\frac{1}{2}$
D. $\frac{4}{\pi}\log_e({2}$
E. $\frac{8}{\pi}$

Answer 1

The general formula for the average value of a function $f(x)$ over $a \le x \le b$ is $$f_{\text{avg}} = \dfrac{1}{b-a}\displaystyle\int_a^bf(x)\,dx$$

Here, you want the avgerage value of the function $f(x) = \tan 2x$ over $0 \le x \le \dfrac{\pi}{8}$.

Can you use the above formula to get the answer?

EDIT: Since you are having trouble integrating $\tan 2x$, notice that $\displaystyle\int \tan 2x\,dx = \int \dfrac{\sin 2x}{\cos 2x}\,dx$.

Now, make a substitution such as $u = \cos 2x$.

Answer 2

$$ \int \tan(2x)\,dx = \frac 1 2\int\frac{1}{\cos(2x)}\Big(\ \underbrace{2\sin(2x)\,dx}_{du}\ \Big) = \frac 1 2 \int \frac 1 u \, du $$ where $u=\cos(2x)$, etc.

Answer 3

it is a good habit to remove irrelevant complications before getting down to detail. here you may observe that the average value of the function $y=\tan(2x)$ over the interval $[0,\frac{\pi}{8}]$ is the same as the average value of $y=\tan x$ over the interval $[0,\frac{\pi}{4}]$. so, using the substitution $w=\tan x$, we obtain: $$ \text{average} = \frac4{\pi}\int_0^{\frac{\pi}{4}}\tan x dx = \frac4{\pi}\int_0^1 \frac{wdw}{1+w^2} = \frac2{\pi}\int_0^1 \frac{2wdw}{1+w^2}=\frac2{\pi}\left[\ln(1+w^2)\right]_0^1=\frac{2 \ln 2}{\pi} $$