How do you find the definite integral $\int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx$?

Hint: This problem deals with finding the integration of the given definite integral. This integration process is done by the method of integration by parts. Integration by parts is done when there is a product of two functions, inside the integral. The integration by parts formula is given below:
\[ \Rightarrow \int {{f_1}(x){f_2}(x)} dx = {f_1}(x)\int {{f_2}(x)} dx - \left[ {\int {f_1^1(x)} \int {{f_2}(x)} dx} \right]dx\]

Complete step-by-step answer:
There are 3 different methods of solving integrations which are:
> Integration by substitution.
> Integration by partial fractions.
> Integration by parts.
Here we are using the method of integration by parts.
Consider the given integral as shown below:
$ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx$
Here let the first function ${f_1}(x) = {x^2}$ and the second function ${f_2}(x) = \cos \left( {\dfrac{1}{5}x} \right)$
Now applying the method of integration by parts to the integral $\int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx$, as shown:
$ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = 5{x^2}\int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {\cos \left( {\dfrac{1}{5}x} \right)} dx - 10\left[ {\int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {\left( {\dfrac{d}{{dx}}\left( {{x^2}} \right)\int {\cos \left( {\dfrac{1}{5}x} \right)} dx} \right)dx} } \right]$
We know that the integration of $\int {\cos xdx = \sin x} $, and the derivative of ${x^2}$ is $2x$, as given below:
$ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = \left[ {{5x^2}\sin \left( {\dfrac{1}{5}x} \right)} \right]_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} - \left[ {\int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {\left( {10x\sin \left( {\dfrac{1}{5}x} \right)} \right)dx} } \right]$
$ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = \left[ {{5x^2}\sin \left( {\dfrac{1}{5}x} \right)} \right]_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} - 10\int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {x\sin \left( {\dfrac{1}{5}x} \right)dx} $
Now applying the integration by parts for the integral $\int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {x\sin \left( {\dfrac{1}{5}x} \right)dx} $, as shown :
$ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = \left[ {{{\left( {\dfrac{{5\pi }}{2}} \right)}^2}\sin \left( {\dfrac{1}{5}\left( {\dfrac{{5\pi }}{2}} \right)} \right) - {{\left( {\dfrac{\pi }{2}} \right)}^2}\sin \left( {\dfrac{1}{5}\left( {\dfrac{\pi }{2}} \right)} \right)} \right] - 2\left[ {x\int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {\sin \left( {\dfrac{1}{5}x} \right)dx - \left[ {\int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {\dfrac{d}{{dx}}\left( x \right)\int {\sin \left( {\dfrac{1}{5}x} \right)} dx} } \right]} dx} \right]$
Substituting the upper and the lower limits of the integral as shown above.
$ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = \left[ {{{\left( {\dfrac{{5\pi }}{2}} \right)}^2}\sin \left( {\dfrac{1}{5}\left( {\dfrac{{5\pi }}{2}} \right)} \right) - {{\left( {\dfrac{\pi }{2}} \right)}^2}\sin \left( {\dfrac{1}{5}\left( {\dfrac{\pi }{2}} \right)} \right)} \right] - 2\left[ {\left[ { - x\cos \left( {\dfrac{1}{5}x} \right)} \right]_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} + \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {1.\cos \left( {\dfrac{1}{5}x} \right)} dx} \right]$
We know that the integration of $\int {\sin xdx = - \cos x} $, hence substituting, as shown above.
\[ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = \left[ {{{\left( {\dfrac{{5\pi }}{2}} \right)}^2}\sin \left( {\dfrac{1}{5}\left( {\dfrac{{5\pi }}{2}} \right)} \right) - {{\left( {\dfrac{\pi }{2}} \right)}^2}\sin \left( {\dfrac{1}{5}\left( {\dfrac{\pi }{2}} \right)} \right)} \right] - 2\left[ { - \left[ {\left( {\dfrac{{5\pi }}{2}} \right)\cos \left( {\dfrac{1}{5}\left( {\dfrac{{5\pi }}{2}} \right)} \right) - \left( {\dfrac{\pi }{2}} \right)\cos \left( {\dfrac{1}{5}\left( {\dfrac{\pi }{2}} \right)} \right)} \right] + \left[ {\sin \left( {\dfrac{1}{5}x} \right)} \right]_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}}} \right]\]
Now we know that the values of $\sin \left( {\dfrac{\pi }{2}} \right) = 1$ and $\cos \left( {\dfrac{\pi }{2}} \right) = 0$, substituting them as shown:
\[ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = \left[ {{{\left( {\dfrac{{5\pi }}{2}} \right)}^2} - {{\left( {\dfrac{\pi }{2}} \right)}^2}\sin \left( {\dfrac{\pi }{{10}}} \right)} \right] - 2\left[ {\left[ {\dfrac{\pi }{2}\cos \left( {\dfrac{\pi }{{10}}} \right)} \right] + \left[ {\sin \left( {\dfrac{\pi }{2}} \right) - \sin \left( {\dfrac{\pi }{{10}}} \right)} \right]} \right]\]
\[ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = \left[ {{{\left( {\dfrac{{5\pi }}{2}} \right)}^2} - {{\left( {\dfrac{\pi }{2}} \right)}^2}\sin \left( {\dfrac{\pi }{{10}}} \right)} \right] + 2\left[ { - \dfrac{\pi }{2}\cos \left( {\dfrac{\pi }{{10}}} \right) - 1 + \sin \left( {\dfrac{\pi }{{10}}} \right)} \right]\]
Now opening the brackets and simplifying the expressions as shown:
\[ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = {\left( {\dfrac{{5\pi }}{2}} \right)^2} - {\left( {\dfrac{\pi }{2}} \right)^2}\sin \left( {\dfrac{\pi }{{10}}} \right) - \pi \cos \left( {\dfrac{\pi }{{10}}} \right) - 2 + 2\sin \left( {\dfrac{\pi }{{10}}} \right)\]
\[ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = \left[ {2 - {{\left( {\dfrac{\pi }{2}} \right)}^2}} \right]\sin \left( {\dfrac{\pi }{{10}}} \right) - \pi \cos \left( {\dfrac{\pi }{{10}}} \right) + {\left( {\dfrac{{5\pi }}{2}} \right)^2} - 2\]
Now substituting the values of \[\sin \left( {\dfrac{\pi }{{10}}} \right) = 0.3090\], \[\cos \left( {\dfrac{\pi }{{10}}} \right) = 0.9510\] and the $\pi = 3.1415$
\[ \Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx = 57.171\]

Final answer: The value of the integral $\int\limits_{\dfrac{\pi }{2}}^{\dfrac{{5\pi }}{2}} {{x^2}\cos \left( {\dfrac{1}{5}x} \right)} dx$ is 57.171.

Note:
Please note that there are two types of integrals, which are definite integrals and indefinite integrals. Definite integrals are those integrals for which there is a particular limit on the integrals. Whereas for the indefinite integrals there are no such limits on the integral.