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Find the area of the following curves:

$$y=7cos(2x) ,quad y=7-7cos(2x) ,quad [0,pi/2].$$

I found the boundaries (calculations not shown but they are correct)

$7cos(2x) > 7-7cos(2x), quad[0,frac{pi}{6}]$

$7-7cos(2x) > 7cos(2x),quad [frac{pi}{6},frac{5pi}{6}]$
begin{align*}
&int 7 cos(2x) - 7-7cos(2x)~dx\
=& frac{7}{2}sin(2x)-7x-frac{7}{2} sin(2x).
end{align*}

begin{align*}
&left[{frac{7}{2}sin(2x)-7x-frac{7}{2} sin(2x)}right]_{0}^{frac{pi}{6}}\
=& -frac{7pi}{2} - 0 =-frac{7pi}{2}.
end{align*}

begin{align*}
&left[7x-frac{7}{2} sin(2x)- frac{7}{2}sin(2x)right]_{frac{pi}{6}}^{frac{5pi}{6}}\
=& frac{35pi}{6}+frac{7sqrt{3}}{2}-frac{7pi}{6}-frac{7sqrt{3}}{2}.
end{align*}

begin{align*}
&left[{frac{7}{2}sin(2x)-7x-frac{7}{2} sin(2x)}right]_{0}^{frac{pi}{6}}+left[7x-frac{7}{2} sin(2x)- frac{7}{2}sin(2x)right]_{frac{pi}{6}}^{frac{5pi}{6}}\
=&-frac{7pi}{2} + frac{35pi}{6}+frac{7sqrt{3}}{2}-frac{7pi}{6}-frac{7sqrt{3}}{2}\
=& frac{7pi}{2}.
end{align*}

This answer was wrong. I double checked my work and have no clue why its not right. To me it seems correct. Can anyone verify?

Check your signs:

$$int 7 cos(2x) - Big(7 - 7cos(2x) Big)$$
$$= int 7 cos(2x) - 7 color{red}{+} 7cos(2x)$$
$$= frac{7}{2} cos(2x) - 7x color{red}{+} frac{7}{2} sin(2x)$$

$$int_0^{pi/6} 14 cos(2x) - 7 dx +int_{pi/6}^{pi/2} 7 - 14 cos(2x) dx$$

$$[7sin(2x) - 7x]_0^{pi/6} + [7x - 7sin(2x)]_{pi/6}^{pi/2}$$

$$[7sin(frac{pi}{3})-frac{7pi}{6}]+[frac{7pi}{2}-7sin(pi)-frac{7pi}{6}+7sin(frac{pi}{3})]$$

$$[7sin(frac{pi}{3})-frac{7pi}{6}]+[frac{7pi}{3}+7sin(frac{pi}{3})]$$

$$ = 2.396986 + 13.392561 = 15.78955$$