Evaluate $\int {{{\cos }^3}xdx} $

Hint: Our integrand is ${\cos ^3}x$ and it can be written as the product of ${\cos ^2}x$ and cos x and using the identity ${\cos ^2}x + {\sin ^2}x = 1$we can split the integral in the form of $I = {I_1} + {I_2}$and using the basic integration values we get the required solution .

Complete step-by-step answer:
Here our integrand is a trigonometric function
${\cos ^3}x$ can be written as the product of ${\cos ^2}x$ and cos x
$ \Rightarrow \int {{{\cos }^3}xdx = \int {\left( {{{\cos }^2}x*\cos x} \right)} dx} $
Now we can use the identity ${\cos ^2}x + {\sin ^2}x = 1$ to replace ${\cos ^2}x$
 $\begin{gathered}
   \Rightarrow \int {{{\cos }^3}xdx = \int {\left( {1 - {{\sin }^2}x} \right)} \cos xdx} \\
   \Rightarrow \int {{{\cos }^3}xdx = \int {\left( {\cos x - \cos x{{\sin }^2}x} \right)} dx} \\
   \Rightarrow \int {{{\cos }^3}xdx = \int {\cos xdx} } - \int {\cos x{{\sin }^2}xdx} \\
\end{gathered} $
So now our integral is of the form $I = {I_1} + {I_2}$
Where ${I_1} = \int {\cos xdx{\text{ , }}{I_2} = \int {\cos x{{\sin }^2}xdx} } $
We know that the integration of cos x is sin x
$ \Rightarrow {I_1} = \sin x + c$
We need to find the value of ${I_2}$ by substitution method
Let sin x = t
Then cos x dx = dt
$ \Rightarrow {I_2} = \int {{t^2}dt} $
We know that $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
Using this
$ \Rightarrow {I_2} = \dfrac{{{t^{2 + 1}}}}{{2 + 1}} + c = \dfrac{{{t^3}}}{3} + c$
And substituting sin x = t
$ \Rightarrow {I_2} = \dfrac{{{{\sin }^3}x}}{3} + c$
Hence
$
   \Rightarrow I = {I_1} + {I_2} \\
   \Rightarrow \int {{{\cos }^3}xdx = \sin x + \dfrac{{{{\sin }^3}x}}{3} + C} \\
$
Hence the required value is obtained .

Note: Many students make a mistake by applying the formula $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $with cos x as x .But it is wrong
And to split ${\cos ^2}x$ students use other identities but it makes the process much tedious.