Evaluate $\int{{{\sin }^{3}}x{{\cos }^{3}}x}$?

Hint: To solve this question, you need to know some basic trigonometric formula and basic integration (by substitution method).
sin$^{2}x+{{\cos }^{2}}x=1$ and $\int{tdt=\dfrac{{{t}^{2}}}{2}}$.(we are going to use both formula in this question).

Complete step by step solution:
Integration: It can be defined as the process of finding the antiderivative. It is used to find many quantities like area, volume etc. integral is of two types
Indefinite integrals: In which lower and upper values are not defined and therefore, we use a constant term in the end.
Definite integration: In which lower and upper values are defined and there is no need to mention a constant term in the end. It gives a finite value.
As given in the question$\int{{{\sin }^{3}}x{{\cos }^{3}}x} dx$ we rearrange the terms and rewrite the integral term as I=$\int{{{\sin }^{3}}x{{\cos }^{2}}x\cos x} dx$ and now we substitute the value of cos$^{2}x$as (sin$^{2}x+{{\cos }^{2}}x=1$ $\Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x$).
After substituting the value of cos$^{2}x$ our integral becomes: I=$\int{{{\sin }^{3}}x}(1-{{\sin }^{2}}x)\cos x$dx after that we substitute the value of sinx = t then on differentiating both sides it becomes cosx dx = t dt.
Integral terms become I = $\int{{{t}^{3}}(1-{{t}^{2}})dt}$ and after simplification it becomes $\int{{{t}^{3}}-{{t}^{5}}dt}$after integration it becomes $\dfrac{{{t}^{4}}}{4}-\dfrac{{{t}^{6}}}{6}+c$( where c is any constant value).
Now we substitute the value of $t$ from above i.e., sinx.
Integral becomes I = $\dfrac{{{\sin }^{4}}x}{4}-\dfrac{{{\sin }^{6}}x}{6}+c$( where c is any constant term).

Note:
You should remember the formula of some basic integration like sinx , cosx , x$^{n}$etc. and know the method of integration by substitution , and after you can easily manipulate the terms and find the result.