Question

Integrate $\int {\sin }^3\left(\mathrm{x}\right){\cos }^3\left(\mathrm{x}\right)\mathrm{dx}$

Answer 1

Integrate the given integral function

Let, $\mathrm{I}=\int {\sin }^3\left(\mathrm{x}\right){\cos }^3\left(\mathrm{x}\right)\mathrm{dx}$

Multiply both sides by $2$

$\Rightarrow \mathrm{I}=\frac{1}{2}\int 2{\sin }^3\left(\mathrm{x}\right){\cos }^3\left(\mathrm{x}\right)\mathrm{dx}$

Let ${\sin }^{2}x=t$

Differentiate both sides,

$$\begin{gathered} \Rightarrow 2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)\mathrm{dx}=\mathrm{dt}...\left(\mathrm{i}\right) \\ \mathrm{I}=\frac{1}{2}\int \mathrm{t}\left(1-\mathrm{t}\right)2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)\mathrm{dx}\left[{\sin }^2\left(\mathrm{x}\right)+{\cos }^2\left(\mathrm{x}\right)=1\right] \end{gathered}$$

By using $\left(\mathrm{i}\right)$,

$$\begin{gathered} \Rightarrow \mathrm{I}=\frac{1}{2}\int \mathrm{t}\left(1-\mathrm{t}\right)\mathrm{dt} \\ \Rightarrow \mathrm{I}=\frac{1}{2}\int \mathrm{tdt}-\frac{1}{2}\int {\mathrm{t}}^2\mathrm{dt}\left[\int {\mathrm{x}}^{\mathrm{n}}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}\right] \\ \Rightarrow \mathrm{I}=\frac{{\mathrm{t}}^2}{4}-\frac{{\mathrm{t}}^3}{6} \\ \Rightarrow \mathrm{I}=\frac{{\sin }^4\left(\mathrm{x}\right)}{4}-\frac{{\sin }^6\left(\mathrm{x}\right)}{6}+\mathrm{C} \end{gathered}$$

Hence, $\int {\sin }^3\left(\mathrm{x}\right){\cos }^3\left(\mathrm{x}\right)\mathrm{dx}$ is integrated as $\frac{{\sin }^4\left(\mathrm{x}\right)}{4}-\frac{{\sin }^6\left(\mathrm{x}\right)}{6}+\mathrm{C}$.