1. ## Trigonometric Calculus

Hi
I am having a bit of trouble with these, so if anyone could help me, I would greatly appreciate it.

1. Find from first principles the derivative of the function f(x) = tan(2x)

2.
a) Find the area between the graph of y = tan x and the x axis for -1≤ x ≤ 1, giving your answer to two decimal places
b) Using calculus, show that ∫ tan x dx = -ln| cos x| + c, where c is an arbitrary constant
c) Find a number k, such that ∫ tan x dx = 2, for 0≤ x≤ k, giving your answer to two decimal places

Thanks again

2. 2. b) $\displaystyle \int{\tan{x}\,dx} = \int{\frac{\sin{x}}{\cos{x}}\,dx}$

$\displaystyle = -\int{\left(\frac{1}{\cos{x}}\right)(-\sin{x})\,dx}$.

Now let $\displaystyle u = \cos{x}$ so that $\displaystyle \frac{du}{dx} = -\sin{x}$ and the integral becomes

$\displaystyle -\int{\left(\frac{1}{u}\right)\,\frac{du}{dx}\,dx}$

$\displaystyle = -\int{\frac{1}{u}\,du}$

$\displaystyle = -\ln{|u|} + C$

$\displaystyle = -\ln{|\cos{x}|} + C$.

3. For 1, I presume you mean using $\displaystyle \lim_{h\to 0}\frac{tan(x+h)- tan(x)}{h}$.

$\displaystyle tan(x)= \frac{sin(x)}{cos(x)}$ so that is $\displaystyle \lim_{h\to 0}\frac{\frac{sin(x+h)}{cos(x+h)}- \frac{sin(x)}{cos(x)}}{h}$$\displaystyle =\lim_{h\to 0}\frac{sin(x+h)cos(x)- sin(x)cos(x+h)}{h cos(x)cos(x+h)}$$\displaystyle = \lim_{h\to 0}\frac{sin(x+h)cos(x)- sin(x)cos(x)+ sin(x)cos(x)- cos(x)sin(x+ h)}{h cos(x)cos(x+h)}$$\displaystyle = \left(cos(x)\lim(h\to 0}\frac{sin(x+h)- sin(x)}{h}- cos(x+h)\lim_{h\to 0}\frac{sin(x+h)- sin(x)}{h}\right)\left(\lim_{h\to 0}\frac{1}{cos(x)cos(x+h)}$

Use limits you know about sine and cosine.

Prove It showed how to do 2b. That should make 2a easy. For 2c, set the formula for $\displaystyle \int_0^k tan(x)dx$, that you get from 2b, equal to 2 and solve for k.