Question

How do you find the integral of $\int{{{\cos }^{3}}x{{\sin }^{2}}xdx}$?

Answer 1

Hint: We first explain the terms $\dfrac{dy}{dx}$ where $y=f\left( x \right)$. We then need to integrate the equation once to find all the solutions of the integration. We take one arbitrary constant term for the integration. We use replacement of base value where $z=\sin x$ for $\int{{{\cos }^{3}}x{{\sin }^{2}}xdx}$. We also use the identity theorem where ${{\cos }^{2}}x+{{\sin }^{2}}x=1$.

Complete step by step solution:
We need to find the integral of $\int{{{\cos }^{3}}x{{\sin }^{2}}xdx}$.
We are going to change the base of the integral where we assume the new variable of $z=\sin x$.
We take the new base and differentiate the equation $z=\sin x$.
We know that the differentiated form of $\sin x$ is $\cos x$.
Differentiating both sides with respect to $x$, we get
 \[\begin{align}
  & \dfrac{d}{dx}\left( z \right)=\dfrac{d}{dx}\left( \sin x \right) \\
 & \Rightarrow \dfrac{dz}{dx}=\cos x \\
\end{align}\]
Now we convert the differentiation into differential form where \[dz=\cos xdx\].
Now we try to reform the main function of the integration where
${{\cos }^{3}}x{{\sin }^{2}}xdx={{\cos }^{2}}x{{\sin }^{2}}x\left( \cos xdx \right)$.
We can also use the identity theorem where ${{\cos }^{2}}x+{{\sin }^{2}}x=1$.
We replace the value for ${{\cos }^{2}}x=1-{{\sin }^{2}}x$.
So, ${{\cos }^{3}}x{{\sin }^{2}}xdx=\left( 1-{{\sin }^{2}}x \right){{\sin }^{2}}x\left( \cos xdx \right)$.
We now replace all those values with \[dz=\cos xdx\] and $z=\sin x$ in the $\int{{{\cos }^{3}}x{{\sin }^{2}}xdx}$
$\begin{align}
  & \int{{{\cos }^{3}}x{{\sin }^{2}}xdx} \\
 & =\int{\left( 1-{{\sin }^{2}}x \right){{\sin }^{2}}x\left( \cos xdx \right)} \\
 & =\int{{{z}^{2}}\left( 1-{{z}^{2}} \right)dz} \\
\end{align}$
We simplify the integral equation more to get
$\int{{{z}^{2}}\left( 1-{{z}^{2}} \right)dz}=\int{\left( {{z}^{2}}-{{z}^{4}} \right)dz}=\int{{{z}^{2}}dz}-\int{{{z}^{4}}dz}$.
Now we use the integral theorem of $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$.
We get $\int{{{z}^{2}}dz}-\int{{{z}^{4}}dz}=\dfrac{{{z}^{3}}}{3}-\dfrac{{{z}^{5}}}{5}+c$.
We put the values where $z=\sin x$.
$\int{{{\cos }^{3}}x{{\sin }^{2}}xdx}=\dfrac{{{\left( \sin x \right)}^{3}}}{3}-\dfrac{{{\left( \sin x \right)}^{5}}}{5}+c=\dfrac{{{\sin }^{3}}x}{3}-\dfrac{{{\sin }^{5}}x}{5}+c$.
The integral of $\int{{{\cos }^{3}}x{{\sin }^{2}}xdx}$ is $\dfrac{{{\sin }^{3}}x}{3}-\dfrac{{{\sin }^{5}}x}{5}+c$. Here $c$ is the integral constant.

Note: We can also solve those integration using the base change for ratio cos. We take $z=\cos x$. In that case the sum gets complicated but the final solution would be the same. It is better to watch out for the odd power value in the ratios and take that as the change in the variable.