How do you find the integral of ${{\cos }^{6}}\left( x \right)$?
Answer
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Hint: To integrate the given function, we first need to convert it into the sum of linear trigonometric functions. For this, we have to use the trigonometric identities; $\cos \left( 3x \right)=4{{\cos }^{3}}\left( x \right)-3\cos \left( x \right)$, $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$, and $\cos \left( 2A \right)=2{{\cos }^{2}}\left( A \right)-1$. By using these identities, the given integral will be converted into the summation of the integrals of the linear trigonometric and the algebraic functions, which can be easily integrated.
Complete step by step answer:
Let us represent the integral given in the question by
$I=\int{{{\cos }^{6}}\left( x \right)}dx$
Clearly, we cannot integrate the given expression directly. This is because only the linear trigonometric functions can be directly integrated. So we need to simplify the given function into the sum of the linear trigonometric functions. Now, we can write the above integral as
$\Rightarrow I=\int{{{\left( {{\cos }^{3}}\left( x \right) \right)}^{2}}}dx........(i)$
Now, we know that
$\cos \left( 3x \right)=4{{\cos }^{3}}\left( x \right)-3\cos \left( x \right)$
Adding $3\cos \left( x \right)$ both the sides, we get
\[\Rightarrow \cos \left( 3x \right)+3\cos \left( x \right)=4{{\cos }^{3}}\left( x \right)\]
Dividing both the sides by $4$, we get
\[\Rightarrow \dfrac{\cos \left( 3x \right)+3\cos \left( x \right)}{4}={{\cos }^{3}}\left( x \right)\]
\[\Rightarrow {{\cos }^{3}}\left( x \right)=\dfrac{1}{4}\cos \left( 3x \right)+\dfrac{3}{4}\cos \left( x \right)..........(ii)\]
Putting the equation (ii) in (i) we get
$\Rightarrow I=\int{{{\left( \dfrac{1}{4}\cos \left( 3x \right)+\dfrac{3}{4}\cos \left( x \right) \right)}^{2}}}dx$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. Considering $a=\dfrac{1}{4}\cos \left( 3x \right)$ and $b=\dfrac{3}{4}\cos \left( x \right)$, we can write the above integral as
\[\Rightarrow I=\int{\left( {{\left( \dfrac{1}{4}\cos \left( 3x \right) \right)}^{2}}+2\left( \dfrac{1}{4}\cos \left( 3x \right) \right)\left( \dfrac{3}{4}\cos \left( x \right) \right)+{{\left( \dfrac{3}{4}\cos \left( x \right) \right)}^{2}} \right)}dx\]
\[\Rightarrow I=\int{\left( \dfrac{1}{16}{{\cos }^{2}}\left( 3x \right)+\dfrac{3}{8}\cos \left( 3x \right)\cos \left( x \right)+\dfrac{9}{16}{{\cos }^{2}}\left( x \right) \right)}dx...........(iii)\]
Now, we know that
$2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$
\[\Rightarrow \cos A\cos B=\dfrac{\cos \left( A+B \right)}{2}+\dfrac{\cos \left( A-B \right)}{2}\]
Putting $A=3x$ and $B=x$, we get
\[\Rightarrow \cos \left( 3x \right)\cos \left( x \right)=\dfrac{\cos \left( 3x+x \right)}{2}+\dfrac{\cos \left( 3x-x \right)}{2}\]
\[\Rightarrow \cos \left( 3x \right)\cos \left( x \right)=\dfrac{\cos \left( 4x \right)}{2}+\dfrac{\cos \left( 2x \right)}{2}.............(iv)\]
Also, we have from the following trigonometric identity
$\cos \left( 2A \right)=2{{\cos }^{2}}\left( A \right)-1$
Adding \[1\] both the sides, we get
\[\Rightarrow \cos \left( 2A \right)+1=2{{\cos }^{2}}\left( A \right)\]
Dividing by $2$ both the sides, we get
\[\Rightarrow \dfrac{\cos \left( 2A \right)+1}{2}={{\cos }^{2}}\left( A \right)\]
\[\Rightarrow {{\cos }^{2}}\left( A \right)=\dfrac{\cos \left( 2A \right)+1}{2}........(v)\]
Putting $A=x$ in (v), we get
\[\Rightarrow {{\cos }^{2}}\left( x \right)=\dfrac{\cos \left( 2x \right)+1}{2}........(vi)\]
Now putting $A=3x$ in (v), we get
\[\Rightarrow {{\cos }^{2}}\left( 3x \right)=\dfrac{\cos \left( 2\left( 3x \right) \right)+1}{2}\]
\[\Rightarrow {{\cos }^{2}}\left( 3x \right)=\dfrac{\cos \left( 6x \right)+1}{2}........(vii)\]
Putting equations (iv), (vi), and (vii) in (iii) we get
\[\begin{align}
& \Rightarrow I=\int{\left( \dfrac{1}{16}\left( \dfrac{\cos \left( 6x \right)+1}{2} \right)+\dfrac{3}{8}\left( \dfrac{\cos \left( 4x \right)}{2}+\dfrac{\cos \left( 2x \right)}{2} \right)+\dfrac{9}{16}\left( \dfrac{\cos \left( 2x \right)+1}{2} \right) \right)}dx \\
& \Rightarrow I=\int{\left( \dfrac{1}{32}\cos \left( 6x \right)+\dfrac{1}{32}+\dfrac{3}{16}\cos \left( 4x \right)+\dfrac{3}{16}\cos \left( 2x \right)+\dfrac{9}{32}\cos \left( 2x \right)+\dfrac{9}{32} \right)}dx \\
\end{align}\]
Writing similar terms together, we get
\[\Rightarrow I=\int{\left( \dfrac{1}{32}\cos \left( 6x \right)+\dfrac{3}{16}\cos \left( 4x \right)+\dfrac{3}{16}\cos \left( 2x \right)+\dfrac{9}{32}\cos \left( 2x \right)+\dfrac{9}{32}+\dfrac{1}{32} \right)}dx\]
On adding the similar terms, we have
\[\begin{align}
& \Rightarrow I=\int{\left( \dfrac{1}{32}\cos \left( 6x \right)+\dfrac{3}{16}\cos \left( 4x \right)+\dfrac{15}{32}\cos \left( 2x \right)+\dfrac{10}{32} \right)}dx \\
& \Rightarrow I=\int{\left( \dfrac{1}{32}\cos \left( 6x \right)+\dfrac{3}{16}\cos \left( 4x \right)+\dfrac{15}{32}\cos \left( 2x \right)+\dfrac{5}{16} \right)}dx \\
\end{align}\]
Now, we can split the integral into four integrals as
\[\Rightarrow I=\int{\left( \dfrac{1}{32}\cos \left( 6x \right) \right)dx+}\int{\left( \dfrac{3}{16}\cos \left( 4x \right) \right)dx}+\int{\dfrac{15}{32}\cos \left( 2x \right)dx}+\int{\dfrac{5}{16}dx}\]
Separating the constants out of all of the integrals, we get
\[\Rightarrow I=\dfrac{1}{32}\int{\cos \left( 6x \right)dx+}\dfrac{3}{16}\int{\cos \left( 4x \right)dx}+\dfrac{15}{32}\int{\cos \left( 2x \right)dx}+\dfrac{5}{16}\int{dx}..........(viii)\]
Now, we know that
\[\int{\cos \theta d\theta }=\sin \theta +C\]
So we have
\[\int{\cos \left( 6x \right)dx}=\dfrac{1}{6}\sin \left( 6x \right)+{{C}_{1}}.........(ix)\]
Also
\[\int{\cos \left( 4x \right)dx}=\dfrac{1}{4}\sin \left( 4x \right)+{{C}_{2}}.........(x)\]
And
\[\int{\cos \left( 2x \right)dx}=\dfrac{1}{2}\sin \left( 2x \right)+{{C}_{3}}.........(xi)\]
Also, we know that
\[\int{dx}=x+{{C}_{4}}..............(xii)\]
Putting (ix), (x), (xi), and (xii) in (viii) we get
\[\begin{align}
& \Rightarrow I=\dfrac{1}{32}\left( \dfrac{1}{6}\sin \left( 6x \right)+{{C}_{1}} \right)+\dfrac{3}{16}\left( \dfrac{1}{4}\sin \left( 4x \right)+{{C}_{2}} \right)+\dfrac{15}{32}\left( \dfrac{1}{2}\sin \left( 2x \right)+{{C}_{3}} \right)+\dfrac{5}{16}\left( x+{{C}_{4}} \right) \\
& \Rightarrow I=\dfrac{1}{192}\sin \left( 6x \right)+\dfrac{1}{32}{{C}_{1}}+\dfrac{3}{64}\sin \left( 4x \right)+\dfrac{3}{16}{{C}_{2}}+\dfrac{15}{64}\sin \left( 2x \right)+\dfrac{15}{32}{{C}_{3}}+\dfrac{5}{16}x+\dfrac{5}{16}{{C}_{4}} \\
& \Rightarrow I=\dfrac{1}{192}\sin \left( 6x \right)+\dfrac{3}{64}\sin \left( 4x \right)+\dfrac{15}{64}\sin \left( 2x \right)+\dfrac{5}{16}x+\dfrac{1}{32}{{C}_{1}}+\dfrac{3}{16}{{C}_{2}}+\dfrac{15}{32}{{C}_{3}}+\dfrac{5}{16}{{C}_{4}} \\
\end{align}\]
Combining all of the constant terms, we finally get
\[\Rightarrow I=\dfrac{1}{192}\sin \left( 6x \right)+\dfrac{3}{64}\sin \left( 4x \right)+\dfrac{15}{64}\sin \left( 2x \right)+\dfrac{5}{16}x+C\]
Hence, the integral of ${{\cos }^{6}}\left( x \right)$ is equal to \[\dfrac{1}{192}\sin \left( 6x \right)+\dfrac{3}{64}\sin \left( 4x \right)+\dfrac{15}{64}\sin \left( 2x \right)+\dfrac{5}{16}x+C\].
Note: Since the integral given in the question is indefinite, so do not forget to add a constant along with the integrated expression. Also, this question can also be solved by using the by-parts method. For that we need to express the given function as the product of $\cos \left( x \right)$ and ${{\cos }^{5}}\left( x \right)$. Taking ${{\cos }^{5}}\left( x \right)$ as the first function, we will be able to obtain the integrated expression. But we must note that the final integrated expression in this case will be different to what we have obtained in the above solution.
Complete step by step answer:
Let us represent the integral given in the question by
$I=\int{{{\cos }^{6}}\left( x \right)}dx$
Clearly, we cannot integrate the given expression directly. This is because only the linear trigonometric functions can be directly integrated. So we need to simplify the given function into the sum of the linear trigonometric functions. Now, we can write the above integral as
$\Rightarrow I=\int{{{\left( {{\cos }^{3}}\left( x \right) \right)}^{2}}}dx........(i)$
Now, we know that
$\cos \left( 3x \right)=4{{\cos }^{3}}\left( x \right)-3\cos \left( x \right)$
Adding $3\cos \left( x \right)$ both the sides, we get
\[\Rightarrow \cos \left( 3x \right)+3\cos \left( x \right)=4{{\cos }^{3}}\left( x \right)\]
Dividing both the sides by $4$, we get
\[\Rightarrow \dfrac{\cos \left( 3x \right)+3\cos \left( x \right)}{4}={{\cos }^{3}}\left( x \right)\]
\[\Rightarrow {{\cos }^{3}}\left( x \right)=\dfrac{1}{4}\cos \left( 3x \right)+\dfrac{3}{4}\cos \left( x \right)..........(ii)\]
Putting the equation (ii) in (i) we get
$\Rightarrow I=\int{{{\left( \dfrac{1}{4}\cos \left( 3x \right)+\dfrac{3}{4}\cos \left( x \right) \right)}^{2}}}dx$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. Considering $a=\dfrac{1}{4}\cos \left( 3x \right)$ and $b=\dfrac{3}{4}\cos \left( x \right)$, we can write the above integral as
\[\Rightarrow I=\int{\left( {{\left( \dfrac{1}{4}\cos \left( 3x \right) \right)}^{2}}+2\left( \dfrac{1}{4}\cos \left( 3x \right) \right)\left( \dfrac{3}{4}\cos \left( x \right) \right)+{{\left( \dfrac{3}{4}\cos \left( x \right) \right)}^{2}} \right)}dx\]
\[\Rightarrow I=\int{\left( \dfrac{1}{16}{{\cos }^{2}}\left( 3x \right)+\dfrac{3}{8}\cos \left( 3x \right)\cos \left( x \right)+\dfrac{9}{16}{{\cos }^{2}}\left( x \right) \right)}dx...........(iii)\]
Now, we know that
$2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$
\[\Rightarrow \cos A\cos B=\dfrac{\cos \left( A+B \right)}{2}+\dfrac{\cos \left( A-B \right)}{2}\]
Putting $A=3x$ and $B=x$, we get
\[\Rightarrow \cos \left( 3x \right)\cos \left( x \right)=\dfrac{\cos \left( 3x+x \right)}{2}+\dfrac{\cos \left( 3x-x \right)}{2}\]
\[\Rightarrow \cos \left( 3x \right)\cos \left( x \right)=\dfrac{\cos \left( 4x \right)}{2}+\dfrac{\cos \left( 2x \right)}{2}.............(iv)\]
Also, we have from the following trigonometric identity
$\cos \left( 2A \right)=2{{\cos }^{2}}\left( A \right)-1$
Adding \[1\] both the sides, we get
\[\Rightarrow \cos \left( 2A \right)+1=2{{\cos }^{2}}\left( A \right)\]
Dividing by $2$ both the sides, we get
\[\Rightarrow \dfrac{\cos \left( 2A \right)+1}{2}={{\cos }^{2}}\left( A \right)\]
\[\Rightarrow {{\cos }^{2}}\left( A \right)=\dfrac{\cos \left( 2A \right)+1}{2}........(v)\]
Putting $A=x$ in (v), we get
\[\Rightarrow {{\cos }^{2}}\left( x \right)=\dfrac{\cos \left( 2x \right)+1}{2}........(vi)\]
Now putting $A=3x$ in (v), we get
\[\Rightarrow {{\cos }^{2}}\left( 3x \right)=\dfrac{\cos \left( 2\left( 3x \right) \right)+1}{2}\]
\[\Rightarrow {{\cos }^{2}}\left( 3x \right)=\dfrac{\cos \left( 6x \right)+1}{2}........(vii)\]
Putting equations (iv), (vi), and (vii) in (iii) we get
\[\begin{align}
& \Rightarrow I=\int{\left( \dfrac{1}{16}\left( \dfrac{\cos \left( 6x \right)+1}{2} \right)+\dfrac{3}{8}\left( \dfrac{\cos \left( 4x \right)}{2}+\dfrac{\cos \left( 2x \right)}{2} \right)+\dfrac{9}{16}\left( \dfrac{\cos \left( 2x \right)+1}{2} \right) \right)}dx \\
& \Rightarrow I=\int{\left( \dfrac{1}{32}\cos \left( 6x \right)+\dfrac{1}{32}+\dfrac{3}{16}\cos \left( 4x \right)+\dfrac{3}{16}\cos \left( 2x \right)+\dfrac{9}{32}\cos \left( 2x \right)+\dfrac{9}{32} \right)}dx \\
\end{align}\]
Writing similar terms together, we get
\[\Rightarrow I=\int{\left( \dfrac{1}{32}\cos \left( 6x \right)+\dfrac{3}{16}\cos \left( 4x \right)+\dfrac{3}{16}\cos \left( 2x \right)+\dfrac{9}{32}\cos \left( 2x \right)+\dfrac{9}{32}+\dfrac{1}{32} \right)}dx\]
On adding the similar terms, we have
\[\begin{align}
& \Rightarrow I=\int{\left( \dfrac{1}{32}\cos \left( 6x \right)+\dfrac{3}{16}\cos \left( 4x \right)+\dfrac{15}{32}\cos \left( 2x \right)+\dfrac{10}{32} \right)}dx \\
& \Rightarrow I=\int{\left( \dfrac{1}{32}\cos \left( 6x \right)+\dfrac{3}{16}\cos \left( 4x \right)+\dfrac{15}{32}\cos \left( 2x \right)+\dfrac{5}{16} \right)}dx \\
\end{align}\]
Now, we can split the integral into four integrals as
\[\Rightarrow I=\int{\left( \dfrac{1}{32}\cos \left( 6x \right) \right)dx+}\int{\left( \dfrac{3}{16}\cos \left( 4x \right) \right)dx}+\int{\dfrac{15}{32}\cos \left( 2x \right)dx}+\int{\dfrac{5}{16}dx}\]
Separating the constants out of all of the integrals, we get
\[\Rightarrow I=\dfrac{1}{32}\int{\cos \left( 6x \right)dx+}\dfrac{3}{16}\int{\cos \left( 4x \right)dx}+\dfrac{15}{32}\int{\cos \left( 2x \right)dx}+\dfrac{5}{16}\int{dx}..........(viii)\]
Now, we know that
\[\int{\cos \theta d\theta }=\sin \theta +C\]
So we have
\[\int{\cos \left( 6x \right)dx}=\dfrac{1}{6}\sin \left( 6x \right)+{{C}_{1}}.........(ix)\]
Also
\[\int{\cos \left( 4x \right)dx}=\dfrac{1}{4}\sin \left( 4x \right)+{{C}_{2}}.........(x)\]
And
\[\int{\cos \left( 2x \right)dx}=\dfrac{1}{2}\sin \left( 2x \right)+{{C}_{3}}.........(xi)\]
Also, we know that
\[\int{dx}=x+{{C}_{4}}..............(xii)\]
Putting (ix), (x), (xi), and (xii) in (viii) we get
\[\begin{align}
& \Rightarrow I=\dfrac{1}{32}\left( \dfrac{1}{6}\sin \left( 6x \right)+{{C}_{1}} \right)+\dfrac{3}{16}\left( \dfrac{1}{4}\sin \left( 4x \right)+{{C}_{2}} \right)+\dfrac{15}{32}\left( \dfrac{1}{2}\sin \left( 2x \right)+{{C}_{3}} \right)+\dfrac{5}{16}\left( x+{{C}_{4}} \right) \\
& \Rightarrow I=\dfrac{1}{192}\sin \left( 6x \right)+\dfrac{1}{32}{{C}_{1}}+\dfrac{3}{64}\sin \left( 4x \right)+\dfrac{3}{16}{{C}_{2}}+\dfrac{15}{64}\sin \left( 2x \right)+\dfrac{15}{32}{{C}_{3}}+\dfrac{5}{16}x+\dfrac{5}{16}{{C}_{4}} \\
& \Rightarrow I=\dfrac{1}{192}\sin \left( 6x \right)+\dfrac{3}{64}\sin \left( 4x \right)+\dfrac{15}{64}\sin \left( 2x \right)+\dfrac{5}{16}x+\dfrac{1}{32}{{C}_{1}}+\dfrac{3}{16}{{C}_{2}}+\dfrac{15}{32}{{C}_{3}}+\dfrac{5}{16}{{C}_{4}} \\
\end{align}\]
Combining all of the constant terms, we finally get
\[\Rightarrow I=\dfrac{1}{192}\sin \left( 6x \right)+\dfrac{3}{64}\sin \left( 4x \right)+\dfrac{15}{64}\sin \left( 2x \right)+\dfrac{5}{16}x+C\]
Hence, the integral of ${{\cos }^{6}}\left( x \right)$ is equal to \[\dfrac{1}{192}\sin \left( 6x \right)+\dfrac{3}{64}\sin \left( 4x \right)+\dfrac{15}{64}\sin \left( 2x \right)+\dfrac{5}{16}x+C\].
Note: Since the integral given in the question is indefinite, so do not forget to add a constant along with the integrated expression. Also, this question can also be solved by using the by-parts method. For that we need to express the given function as the product of $\cos \left( x \right)$ and ${{\cos }^{5}}\left( x \right)$. Taking ${{\cos }^{5}}\left( x \right)$ as the first function, we will be able to obtain the integrated expression. But we must note that the final integrated expression in this case will be different to what we have obtained in the above solution.
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