Question

Use integration by parts to derive the reduction formula \[\int {{{\cos }^n}\left( x \right)dx = \dfrac{1}{n}\sin x{{\cos }^{n - 1}}\left( x \right) + \dfrac{{n - 1}}{n}\int {{{\cos }^{n - 2}}\left( x \right)dx} } \] where \[n\] is a positive integer.
And use the previous reduction formula to evaluate \[\int {{{\cos }^3}x{\text{ }}dx} \]

Answer 1

Hint: To solve this question, we will first split \[{\cos ^n}x\] in to two parts as \[{\cos ^{n - 1}}x\] and \[\cos x\] .Then we will apply the formula of integration by parts i.e., \[\int {\left( {uv} \right)dx = u\int {vdx} - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } {\text{ }}dx\] and derive the given result. After that we will substitute \[n = 3\] in the resultant formula to get the value of \[\int {{{\cos }^3}x{\text{ }}dx} \]

Complete answer:
Let us assume, \[I = \int {{{\cos }^n}x{\text{ }}dx} {\text{ }} - - - \left( A \right)\]
Now let’s split \[{\cos ^n}x\] in to two parts as \[{\cos ^{n - 1}}x\] and \[\cos x\]
i.e., \[{\cos ^n}x = {\cos ^{n - 1}}x \cdot \cos x\]
Therefore, we get
\[I = \int {{{\cos }^{n - 1}}x \cdot \cos x{\text{ }}dx} {\text{ }} - - - \left( i \right)\]
Now we know that
\[\int {\left( {uv} \right)dx = u\int {vdx} - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } {\text{ }}dx\]
So, on applying integration by parts in equation \[\left( i \right)\] we get
\[I = {\cos ^{n - 1}}x\int {\cos x{\text{ }}dx} - \int {\left( {\dfrac{d}{{dx}}\left( {{{\cos }^{n - 1}}x} \right)\int {\cos xdx} } \right)} {\text{ }}dx{\text{ }} - - - \left( {ii} \right)\]
Now we know that
\[\int {\cos x{\text{ }}dx = \sin x} \]
\[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]
Therefore, from equation \[\left( {ii} \right)\] we have
\[I = {\cos ^{n - 1}}x\left( {\sin x} \right) - \int {\left( {n - 1} \right){{\cos }^{n - 2}}x\left( { - \sin x} \right)\sin x{\text{ }}dx} \]
Taking negative sign out from the integral, we get
\[I = {\cos ^{n - 1}}x\left( {\sin x} \right) + \int {\left( {n - 1} \right){{\cos }^{n - 2}}x\left( {\sin x} \right)\sin x{\text{ }}dx} \]
\[\left( {n - 1} \right)\] is a constant term, so we can take it out from the integral
Therefore, we get
\[I = {\cos ^{n - 1}}x\left( {\sin x} \right) + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x\left( {\sin x} \right)\sin x{\text{ }}dx} \]
\[I = {\cos ^{n - 1}}x\left( {\sin x} \right) + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x\left( {{{\sin }^2}x} \right){\text{ }}dx} \]
We know that
\[{\sin ^2}x = 1 - {\cos ^2}x\]
Therefore, we have
\[I = {\cos ^{n - 1}}x\left( {\sin x} \right) + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x\left( {1 - {{\cos }^2}x} \right){\text{ }}dx} \]
On simplifying the integral part, we get
\[I = {\cos ^{n - 1}}x\left( {\sin x} \right) + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x - {{\cos }^n}{\text{x }}dx} \]
\[ \Rightarrow I = {\cos ^{n - 1}}x\left( {\sin x} \right) + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x{\text{ }}dx} - \left( {n - 1} \right)\int {{{\cos }^n}x{\text{ }}dx} \]
Now from equation \[\left( A \right)\] we have \[I = \int {{{\cos }^n}x{\text{ }}dx} \]
Therefore, on substituting the value, we get
\[ \Rightarrow \int {{{\cos }^n}x{\text{ }}dx} = {\cos ^{n - 1}}x\left( {\sin x} \right) + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x{\text{ }}dx} - \left( {n - 1} \right)\int {{{\cos }^n}x{\text{ }}dx} \]
On combining the \[\int {{{\cos }^n}x{\text{ }}dx} \] terms, we get
\[ \Rightarrow \left( {1 + n - 1} \right)\int {{{\cos }^n}x{\text{ }}dx} {\text{ }} = {\cos ^{n - 1}}x\left( {\sin x} \right) + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x{\text{ }}dx} \]
\[ \Rightarrow n\int {{{\cos }^n}x{\text{ }}dx} {\text{ }} = {\cos ^{n - 1}}x\left( {\sin x} \right) + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x{\text{ }}dx} \]
On dividing the above equation by \[n\] we get
\[ \Rightarrow \int {{{\cos }^n}x{\text{ }}dx} {\text{ }} = \dfrac{{{{\cos }^{n - 1}}x\left( {\sin x} \right) + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x{\text{ }}dx} }}{n}\]
\[ \Rightarrow \int {{{\cos }^n}x{\text{ }}dx} {\text{ }} = \dfrac{1}{n}\sin x{\cos ^{n - 1}}x + \dfrac{{n - 1}}{n}\int {{{\cos }^{n - 2}}x{\text{ }}dx} \]
Hence, we get the required result.
Hence, proved
Now we have to evaluate \[\int {{{\cos }^3}x{\text{ }}dx} \]
So, on substituting \[n = 3\] we get
\[ \Rightarrow \int {{{\cos }^3}x{\text{ }}dx} {\text{ }} = \dfrac{1}{3}\sin x{\cos ^{3 - 1}}x + \dfrac{{3 - 1}}{3}\int {{{\cos }^{3 - 2}}x{\text{ }}dx} \]
On simplifying, we get
\[ \Rightarrow \int {{{\cos }^3}x{\text{ }}dx} {\text{ }} = \dfrac{{\sin x{{\cos }^2}x}}{3} + \dfrac{2}{3}\int {\cos x{\text{ }}dx} \]
We know that
\[\int {\cos x{\text{ }}dx = \sin x} \]
Therefore, we get
\[ \Rightarrow \int {{{\cos }^3}x{\text{ }}dx} {\text{ }} = \dfrac{{\sin x{{\cos }^2}x}}{3} + \dfrac{2}{3}\sin x\]
We know that
\[{\cos ^2}x = 1 - {\sin ^2}x\]
Therefore, we have
\[ \Rightarrow \int {{{\cos }^3}x{\text{ }}dx} {\text{ }} = \dfrac{{\sin x\left( {1 - {{\sin }^2}x} \right)}}{3} + \dfrac{2}{3}\sin x\]
\[ \Rightarrow \int {{{\cos }^3}x{\text{ }}dx} {\text{ }} = \dfrac{{\sin x - {{\sin }^3}x}}{3} + \dfrac{2}{3}\sin x\]
On taking L.C.M we get
\[ \Rightarrow \int {{{\cos }^3}x{\text{ }}dx} {\text{ }} = \dfrac{{\sin x - {{\sin }^3}x + 2\sin x}}{3}\]
\[ \Rightarrow \int {{{\cos }^3}x{\text{ }}dx} {\text{ }} = \dfrac{{3\sin x - {{\sin }^3}x}}{3}\]
On dividing by \[3\] we get
\[ \Rightarrow \int {{{\cos }^3}x{\text{ }}dx} {\text{ }} = \sin x - \dfrac{{{{\sin }^3}x}}{3}\]
As it is an indefinite integral, so add constant of integration.
Hence, we get the final result as
\[ \Rightarrow \int {{{\cos }^3}x{\text{ }}dx} {\text{ }} = \sin x - \dfrac{{{{\sin }^3}x}}{3} + c\]

Note: While solving this question, keep track of each step as the solution involves complex calculations, so there is a high probability of error. Also make sure you know the differentiation and integration of \[\cos x\] as students get confused between the two. The differentiation of \[\cos x\] is \[ - \sin x\] while the integration of \[\cos x\] is \[\sin x\] . So be aware of each and every formula as one mistake can lead you to the wrong answer.