Question

What is the integral of $\int{{{\sin }^{3}}}x{{\cos }^{4}}xdx$ .

Answer 1

Hint: To find the integral of $\int{{{\sin }^{3}}}x{{\cos }^{4}}xdx$ , we have to equate $t=\cos x$ . We will have to differentiate t with respect to x. Then from this, we have to find dx. We should substitute these values in the given integral. Then, we have to use the trigonometric identity ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ . Then we have to simplify and integrate and substitute back for t.

Complete step by step solution:
We have to find the integral of $\int{{{\sin }^{3}}}x{{\cos }^{4}}xdx$ . Let us denote $t=\cos x...\left( i \right)$ .
We have to differentiate t with respect to x.
$\begin{align}
  & \Rightarrow \dfrac{dt}{dx}=-\sin x \\
 & \Rightarrow dt=-\sin xdx \\
\end{align}$
From the above, we have to find dx.
$\Rightarrow dx=-\dfrac{dt}{\sin x}...\left( ii \right)$
Let us substitute (i) and (ii) in the given integral.
$\Rightarrow \int{{{\sin }^{3}}}x{{\cos }^{4}}xdx=\int{{{\sin }^{3}}x\times {{t}^{4}}\times -\dfrac{dt}{\sin x}}$
Let us cancel sin x from the numerator and denominator.
$\Rightarrow \int{{{\sin }^{3}}}x{{\cos }^{4}}xdx=\int{-{{\sin }^{2}}x\times {{t}^{4}}\times dt}...\left( iii \right)$
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ .
$\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x$
Let us substitute the above identity in equation (iii).
 $\Rightarrow \int{{{\sin }^{3}}}x{{\cos }^{4}}xdx=\int{-\left( 1-{{\cos }^{2}}x \right)\times {{t}^{4}}\times dt}...\left( iii \right)$
Let us substitute (i) in the above equation.
$\Rightarrow \int{{{\sin }^{3}}}x{{\cos }^{4}}xdx=\int{-\left( 1-{{t}^{2}} \right){{t}^{4}}dt}$
Let us apply distributive law.
$\Rightarrow \int{{{\sin }^{3}}}x{{\cos }^{4}}xdx=\int{\left( -{{t}^{4}}+{{t}^{6}} \right)dt}$
Now, we have to apply the property $\int{\left( mx+nx \right)dx}=\int{mxdx}+\int{nxdx}$ . Then, the above equation can be written as
$\Rightarrow \int{{{\sin }^{3}}}x{{\cos }^{4}}xdx=\int{-{{t}^{4}}dt}+\int{{{t}^{6}}dt}$
We know that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$ . The integral of the above equation becomes
$\begin{align}
  & \Rightarrow \int{{{\sin }^{3}}}x{{\cos }^{4}}xdx=-\dfrac{{{t}^{4+1}}}{4+1}+\dfrac{{{t}^{6+1}}}{6+1}+C \\
 & \Rightarrow \int{{{\sin }^{3}}}x{{\cos }^{4}}xdx=-\dfrac{1}{5}{{t}^{5}}+\dfrac{1}{7}{{t}^{7}}+C \\
\end{align}$
Now, we have to substitute back the value of t from equation (i). Then, the above equation becomes
$\Rightarrow \int{{{\sin }^{3}}}x{{\cos }^{4}}xdx=-\dfrac{1}{5}{{\cos }^{5}}x+\dfrac{1}{7}{{\cos }^{7}}x+C$

Hence, the integral of $\int{{{\sin }^{3}}}x{{\cos }^{4}}xdx$ is $-\dfrac{1}{5}{{\cos }^{5}}x+\dfrac{1}{7}{{\cos }^{7}}x+C$

Note: Students must know how to integrate and integrate basic functions. They must also know the trigonometric properties. Students must never forget to put the constant after integration. They must not forget to substitute back the value of t.