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# ?

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Answer 1 · 2014-09-25T00:14:36

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(x)=2sin(x)#

#tan(x)-2sin(x)=0#

#sin(x)/cos(x)-2sin(x)=0#

Factor out #sin(x)#

#sin(x)(1/cos(x)-2)=0#

Set each factor equal to zero to find the intersections.

#sin(x)=0#

#x=0#

#1/cos(x)-2=0#

#1/cos(x)=2#

#2cos(x)=1#

#cos(x)=1/2#

#x=(-pi)/3,pi/3#

The functions intersect at #-pi/3,0, and pi/3#

Our 2 new intervals are #[-pi/3,0]# and #[0,pi/3]#.

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

For the first interval , #[-pi/3,0]#, lets use #-pi/4# because it falls within the interval. Let's test that value with both functions.

#tan(-pi/4)=1# Greater

#2sin(-pi/4)<0#

So the first integral looks like, #int_(-pi/3)^0tan(x)-2sin(x)dx#

For the second interval , #[0,pi/3]#, lets use #pi/4# because it falls within the interval. Let's test that value with both functions.

#tan(pi/4)=1#

#2sin(pi/4)=2# Greater

So the first integral looks like, #int_0^(pi/3) 2sin(x)-tan(x)dx#

We now have to solve the following ...

#int_(-pi/3)^0tan(x)-2sin(x)dx# + #int_0^(pi/3) 2sin(x)-tan(x)dx#

#=[ln |sec (x)|]_(-pi/3)^0-2*[-cos(x)]_(-pi/3)^0+2*[-cos(x)]_0^(pi/3)-[ln|sec(x)|]_0^(pi/3)#

#=[ln |sec (0)|-ln |sec (-pi/3)|]-2*[-cos(0)-(-cos(-pi/3))]+2*[-cos(pi/3)-(-cos(0))]-[ln|sec(pi/3)|-ln|sec(0)|]#

Remember that #sec(x) = 1/cos(x)# when inputting this into a calculator.

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

#=0.6137056389#