Question

30-32 sketch the region enclosed by the given curves and find its area.

Answer 1

Check the picture below.

so hmmm for the sake of completion, let's check where they intersect

$cos^2(x)sin(x)=sin(x)\implies cos^2(x)sin(x)-sin(x)=0 \\\\\\ sin(x)[cos^2(x)-1]=0\implies \begin{cases} sin(x)=0\\ cos^2(x)-1=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ sin(x)=0\implies x=sin^{-1}(0)\implies \boxed{x=0} \\\\[-0.35em] ~\dotfill\\\\ cos^2(x)-1=0\implies cos^2(x)=1\implies cos(x)=\sqrt{1} \\\\\\ cos(x)=1\implies x=cos^{-1}(1)\implies \boxed{x=\pi}$

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.

$\stackrel{above}{sin(x)}~~ - ~~\stackrel{below}{cos^2(x)sin(x)} \implies sin(x)-[1-sin^2(x)]sin(x) \\\\\\ sin(x)-[sin(x)-sin^3(x)]\implies ~~\begin{matrix} sin(x)-sin(x) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ +sin^3(x)\implies sin^3(x)$

now let's use the triple angle identity of sine

$\stackrel{\textit{triple angle identity}}{sin(3x)=3sin(x)-4sin^3(x)}\implies 4sin^3(x)=3sin(x)-sin(3x) \\\\\\ sin^3(x)=\cfrac{3sin(x)-sin(3x)}{4}\implies sin^3(x)=\cfrac{3}{4}sin(x)-\cfrac{1}{4}sin(3x) \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int\limits_{0}^{\pi }~sin^3(x)dx\implies \int\limits_{0}^{\pi }~\left[ \cfrac{3}{4}sin(x)-\cfrac{1}{4}sin(3x) \right]dx$

$\displaystyle \cfrac{3}{4}\int\limits_{0}^{\pi }sin(x)dx-\cfrac{1}{4}\int\limits_{0}^{\pi }sin(3x)dx\implies \left. \cfrac{3}{4} \cdot -cos(x) \right]_{0}^{\pi }-\left. \cfrac{1}{4} \cdot \cfrac{-cos(3x)}{3} \right]_{0}^{\pi } \\\\\\ \cfrac{3}{2}~~ - ~~\cfrac{1}{6}\implies \implies \blacktriangleright \cfrac{4}{3} \blacktriangleleft$