Answer:
Area (A) [tex]\simeq[/tex] 0.0682
Step-by-step explanation:
The sketch for the region enclosed by the given curves can be found in the image attached below.
From the image below;
The two curves intersect in the area of tan 9x = 2 sin 9x
Recall that:
[tex]\dfrac{sin \ 9x }{cos \ 9x } = 2 \ sin \ 9x[/tex]
[tex]cos \ 9 x= \dfrac{1}{2}[/tex]
making x the subject; then:
[tex]x = \dfrac{1}{9} cos ^{-1}(\dfrac{1}{2})[/tex]
[tex]x = \dfrac{1}{9}(\dfrac{\pi}{3})[/tex]
[tex]x = \dfrac{\pi}{27}[/tex]
The subdivision of the domain is in two intervals [tex][-\dfrac{\pi}{27}, 0], [0, \dfrac{\pi}{27}][/tex]
where;
[tex]x \ \ \varepsilon \ \ [ -\dfrac{\pi}{27},0]; tan \ 9x > 2 sin \ 9x[/tex]
[tex]x \ \ \varepsilon \ \ [ 0,\dfrac{\pi}{27}]; tan \ 9x < 2 \ sin \ 9x[/tex]
Area (A) = [tex]\int ^0_{-\pi/27}}} ( tan \ 9x - 2 \ sin \ 9x ) \ dx + \int ^{\pi/27}_{0}(2 \ sin \ 9x - tan \ 9x \ ) \ dx[/tex]
[tex]= \begin {bmatrix} \dfrac{1}{9} In \ sec 9x + \dfrac{2}{9} \ cos \ 9x \end {bmatrix}^0_{-\pi/27} + \begin {bmatrix} -\dfrac{2}{7} cos 9x - \dfrac{1}{9} \ In \ sec \ 9x \end {bmatrix}^{-\pi/27} _0[/tex]
[tex]= \dfrac{1}{9}(1-In 2) + \dfrac{1}{9}(1-In 2)\\[/tex]
[tex]= \dfrac{2}{9}(1 - In 2)[/tex]
Area (A) = 0.06818
Area (A) [tex]\simeq[/tex] 0.0682
