${\int }_0^{\pi } {\cos }^{3}xdx$ is equal to 

Let

 $\begin{array}{l}I={\int }_0^{\pi } {\cos }^{3}xdx\\ ={\int }_0^{\pi } \left(1-{\sin }^{2}x\right)\cos xdx\end{array}$

Put sin x = t $\Rightarrow$ cos x dx = dt Now  if x =0 then t = 0 

                                                         if x=$\mathrm{\pi }$  then t=0
So, $\mathrm{I}={\int }_0^0 \left(1-{\mathrm{t}}^2\right)\mathrm{dt}=\mathrm{t}-{\left.\frac{{\mathrm{t}}^3}{3}\right|}_0^0=0$

Remember: The integral range bound between 0 and 0 will always gives the result as zero.