What is the area bounded by the two graphs $y =\sin x$ and $y =\sin x\cos2x$? $(0\le x\le\pi)$ What is the area bounded by the two graphs $y =\sin x$ and $y =\sin x\cos2x$? $(0\le x\le\pi)$


My approach

$$\int^\pi_0\sin x -\sin x \cos2x \ dx$$
$$\int^\pi_0\sin x (1-\cos2x) \ dx$$

As $\cos2x = \cos(x+x)$

$\cos2x =\cos x \cos x -\sin x \sin x =\cos^2x-\sin^2x$

And as $\sin^2x + \cos^2x=1,$ 

$$\int^\pi_0\sin x (1-\cos2x) \ dx = \int^\pi_0\sin x \left(\sin^2x + \cos^2x-(\cos^2x-\sin^2x)\right) \ dx$$

$$=\int^\pi_0\sin x (2\sin^2x) = \int^\pi_02\sin^3x \ dx$$

Doing the integration gives me

$$\left[\frac{\sin^4x}{2\cos x}\right]^\pi_0$$
which is 0. Using technology to verify this answer, it was $\frac{8}{3}$ instead of 0. A calculator suggests my approach is valid, but there is clearly something wrong in my calculation.
Once you arrived to
$$=\int^\pi_0\sin x (2\sin^2x) dx$$
you do the following
$$\int_0^{\pi } 2 \sin (x) \left(1-\cos ^2(x)\right) \, dx$$
and then substitute $\cos(x)=u\rightarrow -\sin(x)\,dx=du$
The integral becomes
$$\int_1^{-1} -2 \left(1-u^2\right) \, du=\left. 2 \left(\frac{u^3}{3}-u\right)\right|_1^{-1}=\frac{8}{3}$$
Cause your last integral is incorrect, $$ \int^\pi_02\sin^3x \ dx = \left[\frac{1}{12} (\cos(3 x) - 9\cos(x))\right]_0^{\pi}  \ne \left[\frac{\sin^4x}{2\cos x}\right]^\pi_0$$