
How do you integrate \[\int{{{\cos }^{3}}\left( \dfrac{x}{3} \right)}\] ?
Answer
548.7k+ views
Hint: In order to solve the above question, we have to apply trigonometric substitutions. First, we will split the cos function using trigonometric identities and make substitutions according to that. After that we will integrate term by term using simple integration formulas.
Complete step by step answer:
The above question belongs to the concept of integration by trigonometric substitution. Here we have to use basic trigonometric substitutions in order to integrate the given function. We have to integrate \[\int{{{\cos }^{3}}\left( \dfrac{x}{3} \right)}\].
We will first make a few substitutions.
Our first step is to let \[t=\dfrac{x}{3}\]
\[\Rightarrow dt=\dfrac{1}{3}dx\]
Now replacing t in the given integral.
\[\int{{{\cos }^{3}}\left( \dfrac{x}{3} \right)}=3\int{{{\cos }^{3}}\left( \dfrac{x}{3} \right)\dfrac{1}{3}=3}\int{{{\cos }^{3}}\left( t \right)dt}\]
As we don’t have any direct integral for this, so we will first split the \[{{\cos }^{3}}t\] function into \[{{\cos }^{2}}t\] and \[\cos t\] . After that we will rewrite the \[{{\cos }^{2}}t\] using the trigonometric identity \[{{\cos }^{2}}t+{{\sin }^{2}}t=1\] .
\[\Rightarrow {{\cos }^{2}}t=1-{{\sin }^{2}}t\]
Therefore, the integral becomes.
\[\begin{align}
& 3\int{{{\cos }^{3}}\left( t \right)dt}=3\int{{{\cos }^{2}}\left( t \right)\cos (t)dt} \\
& \Rightarrow 3\int{{{\cos }^{3}}\left( t \right)dt}=3\int{(1-{{\sin }^{2}}t)\cos tdt} \\
& \Rightarrow 3\int{{{\cos }^{3}}\left( t \right)dt}=3\int{1-{{s}^{2}}}ds \\
\end{align}\]
Where sin t is substituted with s, therefore ds= cos t dt.
Now we will integrate term by term using the integrating rule \[\int{{{s}^{n}}ds=\dfrac{{{s}^{n+1}}}{n+1}}\] ,here n value should not be equal to -1.
After applying the integration rule, we get.
\[\begin{align}
& 3\int{1-{{s}^{2}}}ds=3\left( s-\dfrac{{{s}^{3}}}{3} \right) \\
& \Rightarrow 3\int{1-{{s}^{2}}}ds=3s-{{s}^{3}} \\
\end{align}\]
Now reverse the substitution which we used \[s=\sin t,t=\dfrac{x}{3}\]
\[3s-{{s}^{2}}=3\sin \left( \dfrac{x}{3} \right)-{{\sin }^{3}}\left( \dfrac{x}{3} \right)+c\]
Therefore, the integration of the given integral \[\int{{{\cos }^{3}}\left( \dfrac{x}{3} \right)}\] is \[3\sin \left( \dfrac{x}{3} \right)-{{\sin }^{3}}\left( \dfrac{x}{3} \right)+c\].
Note: While solving the above question be careful with the integration part. Do remember the substitution method used here for future use. Try to solve the question step by step. To solve these types of questions, we should have the knowledge of trigonometric identities. Do not forget to reverse the substitution. After integration add the constant of integration in the result.
Complete step by step answer:
The above question belongs to the concept of integration by trigonometric substitution. Here we have to use basic trigonometric substitutions in order to integrate the given function. We have to integrate \[\int{{{\cos }^{3}}\left( \dfrac{x}{3} \right)}\].
We will first make a few substitutions.
Our first step is to let \[t=\dfrac{x}{3}\]
\[\Rightarrow dt=\dfrac{1}{3}dx\]
Now replacing t in the given integral.
\[\int{{{\cos }^{3}}\left( \dfrac{x}{3} \right)}=3\int{{{\cos }^{3}}\left( \dfrac{x}{3} \right)\dfrac{1}{3}=3}\int{{{\cos }^{3}}\left( t \right)dt}\]
As we don’t have any direct integral for this, so we will first split the \[{{\cos }^{3}}t\] function into \[{{\cos }^{2}}t\] and \[\cos t\] . After that we will rewrite the \[{{\cos }^{2}}t\] using the trigonometric identity \[{{\cos }^{2}}t+{{\sin }^{2}}t=1\] .
\[\Rightarrow {{\cos }^{2}}t=1-{{\sin }^{2}}t\]
Therefore, the integral becomes.
\[\begin{align}
& 3\int{{{\cos }^{3}}\left( t \right)dt}=3\int{{{\cos }^{2}}\left( t \right)\cos (t)dt} \\
& \Rightarrow 3\int{{{\cos }^{3}}\left( t \right)dt}=3\int{(1-{{\sin }^{2}}t)\cos tdt} \\
& \Rightarrow 3\int{{{\cos }^{3}}\left( t \right)dt}=3\int{1-{{s}^{2}}}ds \\
\end{align}\]
Where sin t is substituted with s, therefore ds= cos t dt.
Now we will integrate term by term using the integrating rule \[\int{{{s}^{n}}ds=\dfrac{{{s}^{n+1}}}{n+1}}\] ,here n value should not be equal to -1.
After applying the integration rule, we get.
\[\begin{align}
& 3\int{1-{{s}^{2}}}ds=3\left( s-\dfrac{{{s}^{3}}}{3} \right) \\
& \Rightarrow 3\int{1-{{s}^{2}}}ds=3s-{{s}^{3}} \\
\end{align}\]
Now reverse the substitution which we used \[s=\sin t,t=\dfrac{x}{3}\]
\[3s-{{s}^{2}}=3\sin \left( \dfrac{x}{3} \right)-{{\sin }^{3}}\left( \dfrac{x}{3} \right)+c\]
Therefore, the integration of the given integral \[\int{{{\cos }^{3}}\left( \dfrac{x}{3} \right)}\] is \[3\sin \left( \dfrac{x}{3} \right)-{{\sin }^{3}}\left( \dfrac{x}{3} \right)+c\].
Note: While solving the above question be careful with the integration part. Do remember the substitution method used here for future use. Try to solve the question step by step. To solve these types of questions, we should have the knowledge of trigonometric identities. Do not forget to reverse the substitution. After integration add the constant of integration in the result.
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