# How do you find the area of the region bounded by the curves y=tan(x) and y=2sin(x) on the interval -pi/3<=x<=pi/3 ?

Sep 25, 2014

We have a bit of work ahead of use with this problem. This is a definite integral problem.

First we need to identify the location(s) where these functions intersect. These will be the x-values and also define the upper and lower boundaries.

$\tan \left(x\right) = 2 \sin \left(x\right)$

$\tan \left(x\right) - 2 \sin \left(x\right) = 0$

$\sin \frac{x}{\cos} \left(x\right) - 2 \sin \left(x\right) = 0$

Factor out $\sin \left(x\right)$

$\sin \left(x\right) \left(\frac{1}{\cos} \left(x\right) - 2\right) = 0$

Set each factor equal to zero to find the intersections.

$\sin \left(x\right) = 0$

$x = 0$

$\frac{1}{\cos} \left(x\right) - 2 = 0$

$\frac{1}{\cos} \left(x\right) = 2$

$2 \cos \left(x\right) = 1$

$\cos \left(x\right) = \frac{1}{2}$

$x = \frac{- \pi}{3} , \frac{\pi}{3}$

The functions intersect at $- \frac{\pi}{3} , 0 , \mathmr{and} \frac{\pi}{3}$

Our 2 new intervals are $\left[- \frac{\pi}{3} , 0\right]$ and $\left[0 , \frac{\pi}{3}\right]$.

Now we need to figure out which function is greater over each of the 2 new intervals.

For the first interval , $\left[- \frac{\pi}{3} , 0\right]$, lets use $- \frac{\pi}{4}$ because it falls within the interval. Let's test that value with both functions.

$\tan \left(- \frac{\pi}{4}\right) = 1$ Greater

$2 \sin \left(- \frac{\pi}{4}\right) < 0$

So the first integral looks like, ${\int}_{- \frac{\pi}{3}}^{0} \tan \left(x\right) - 2 \sin \left(x\right) \mathrm{dx}$

For the second interval , $\left[0 , \frac{\pi}{3}\right]$, lets use $\frac{\pi}{4}$ because it falls within the interval. Let's test that value with both functions.

$\tan \left(\frac{\pi}{4}\right) = 1$

$2 \sin \left(\frac{\pi}{4}\right) = 2$ Greater

So the first integral looks like, ${\int}_{0}^{\frac{\pi}{3}} 2 \sin \left(x\right) - \tan \left(x\right) \mathrm{dx}$

We now have to solve the following ...

${\int}_{- \frac{\pi}{3}}^{0} \tan \left(x\right) - 2 \sin \left(x\right) \mathrm{dx}$ + ${\int}_{0}^{\frac{\pi}{3}} 2 \sin \left(x\right) - \tan \left(x\right) \mathrm{dx}$

$= {\left[\ln | \sec \left(x\right) |\right]}_{- \frac{\pi}{3}}^{0} - 2 \cdot {\left[- \cos \left(x\right)\right]}_{- \frac{\pi}{3}}^{0} + 2 \cdot {\left[- \cos \left(x\right)\right]}_{0}^{\frac{\pi}{3}} - {\left[\ln | \sec \left(x\right) |\right]}_{0}^{\frac{\pi}{3}}$

$= \left[\ln | \sec \left(0\right) | - \ln | \sec \left(- \frac{\pi}{3}\right) |\right] - 2 \cdot \left[- \cos \left(0\right) - \left(- \cos \left(- \frac{\pi}{3}\right)\right)\right] + 2 \cdot \left[- \cos \left(\frac{\pi}{3}\right) - \left(- \cos \left(0\right)\right)\right] - \left[\ln | \sec \left(\frac{\pi}{3}\right) | - \ln | \sec \left(0\right) |\right]$

Remember that $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ when inputting this into a calculator.

After entering this huge expression into the TI-84 C the results is as follows ...

$= 0.6137056389$