Integration: find the shaded area in diagram bound by the curves $y=\sin(2x)$ and $y=\cos(x)$ for: $\frac{-\pi}{2} \le x \le \frac{\pi}{2}$

Question

Find the shaded area bound by the curves $y=\sin(2x)$ and $y=\cos(x)$ for:

$$\frac{-\pi}{2} \le x \le \frac{\pi}{2}$$

What I have tried so far:

For part (a):

area under $y=\cos(x)$ is given by

$$\int_{\pi/2}^{-\pi/2} \cos(x)dx=-2⇒2 \,\text{units}^2$$

area under $y=\sin(2x)$ is given by

$$\int_{0}^{-\pi/2} \sin(2x)dx=1$$

$$\int_{\pi/2}^{0} \sin(2x)dx=-1$$

total = $2\, \text{units}^2$

Would the fact that they are the same area, be helpful in finding the solution? So both areas under the curves for the given domain $\frac{-\pi}{2} \le x \le\frac{\pi}{2}$ are $2\, \text{units}^2$, but I'm still unsure how to proceed from here for both part (a) and (b).

Any hints would be much appreciated!

This is from a Year 12 Methods book.

Answer 1

The two curves cross at $x=-\pi/6$, so the shades area id given as $$A=\int_{-\pi/2}^{\pi/6} [\cos x-\sin 2x] dx+\int _{\pi/6}^{\pi/2} [\sin 2x -\cos x] dx=\frac{9}{4}+\frac{1}{4}=\frac{5}{2}.$$