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

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.