Check the picture below.
so hmmm for the sake of completion, let's check where they intersect
![cos^2(x)sin(x)=sin(x)\implies cos^2(x)sin(x)-sin(x)=0 \\\\\\ sin(x)[cos^2(x)-1]=0\implies \begin{cases} sin(x)=0\\ cos^2(x)-1=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ sin(x)=0\implies x=sin^(-1)(0)\implies \boxed{x=0} \\\\[-0.35em] ~\dotfill\\\\ cos^2(x)-1=0\implies cos^2(x)=1\implies cos(x)=√(1) \\\\\\ cos(x)=1\implies x=cos^(-1)(1)\implies \boxed{x=\pi}](https://img.qammunity.org/2023/formulas/mathematics/high-school/9y7p6wvlxwebo1ionbj3361x9q7yxvxtgi.png)
now, notice, in the picture the function that is "above" or the "ceiling" function is the sin(x), so we'll get the area under the curve by using "above" - "below" functions.
![\stackrel{above}{sin(x)}~~ - ~~\stackrel{below}{cos^2(x)sin(x)} \implies sin(x)-[1-sin^2(x)]sin(x) \\\\\\ sin(x)-[sin(x)-sin^3(x)]\implies ~~\begin{matrix} sin(x)-sin(x) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ +sin^3(x)\implies sin^3(x)](https://img.qammunity.org/2023/formulas/mathematics/high-school/59deaty3ov4qwaw8460nlmfjh4zwa5yt48.png)
now let's use the triple angle identity of sine
![\stackrel{\textit{triple angle identity}}{sin(3x)=3sin(x)-4sin^3(x)}\implies 4sin^3(x)=3sin(x)-sin(3x) \\\\\\ sin^3(x)=\cfrac{3sin(x)-sin(3x)}{4}\implies sin^3(x)=\cfrac{3}{4}sin(x)-\cfrac{1}{4}sin(3x) \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int\limits_(0)^(\pi )~sin^3(x)dx\implies \int\limits_(0)^(\pi )~\left[ \cfrac{3}{4}sin(x)-\cfrac{1}{4}sin(3x) \right]dx](https://img.qammunity.org/2023/formulas/mathematics/high-school/wfsiajo7eocuifallwc2r4jrbhb2nptptl.png)
![\displaystyle \cfrac{3}{4}\int\limits_(0)^(\pi )sin(x)dx-\cfrac{1}{4}\int\limits_(0)^(\pi )sin(3x)dx\implies \left. \cfrac{3}{4} \cdot -cos(x) \right]_(0)^(\pi )-\left. \cfrac{1}{4} \cdot \cfrac{-cos(3x)}{3} \right]_(0)^(\pi ) \\\\\\ \cfrac{3}{2}~~ - ~~\cfrac{1}{6}\implies \implies \blacktriangleright \cfrac{4}{3} \blacktriangleleft](https://img.qammunity.org/2023/formulas/mathematics/high-school/7gqr9y7j4rnvkpoi882oksy32zdouvvz1b.png)