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Question: How do you find the definite integral \[\int\limits_{\dfrac{\pi }{2}}^{\dfrac{5\pi }{2}}{{{x}^{2}}\c...

How do you find the definite integral π25π2x2cos(15x)dx\int\limits_{\dfrac{\pi }{2}}^{\dfrac{5\pi }{2}}{{{x}^{2}}\cos \left( \dfrac{1}{5}x \right)}dx?

Explanation

Solution

In this problem, we have to find the value of the given definite integral. We can first find the indefinite integral by using the u-v method formula, which is udv=uvvdu\int{udv=uv-\int{vdu}} and then we can use the given limits i.e. in the definite integral. we can then subtract the definite integral at the lower limit from the definite integral at the upper limit and we can calculate it to get the value.

Complete step-by-step solution:
We know that the given definite integral is,
π25π2x2cos(15x)dx\Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\dfrac{5\pi }{2}}{{{x}^{2}}\cos \left( \dfrac{1}{5}x \right)}dx…….. (1)
We can now write the indefinite integral of the above integral, we get
x2cos(15x)dx\Rightarrow \int{{{x}^{2}}\cos \left( \dfrac{1}{5}x \right)dx}………. (2)
We know that, by analysing the above integral, we can use the integration by parts formula to integrate it.
We know that the integration by parts formula is,
udv=uvvdu\Rightarrow \int{udv=uv-\int{vdu}} ……… (3)
We can now use the ILATE rule, where we have a product of two functions.
In this rule, we can take the left term as the first function and the right term as the second function.
Here we can take the first term as the first function in such a way that the first function could be easily integrated.
We can now take the above indefinite integral (2).
Let u=x2u={{x}^{2}} and v=5sin(15x)v=5\sin \left( \dfrac{1}{5}x \right)
Then, du=2xdxdu=2xdx and dv=cos(15x)dv=\cos \left( \dfrac{1}{5}x \right) .
We can substitute the above values in the formula (3), we get
x2cos(15x)dx=5x2sin(15x)10xsin(15x)dx\Rightarrow \int{{{x}^{2}}\cos \left( \dfrac{1}{5}x \right)dx=5{{x}^{2}}\sin \left( \dfrac{1}{5}x \right)-10\int{x\sin \left( \dfrac{1}{5}x \right)dx}} ……. (4)
We can now use the integration by part in integral 10xsin(15x)dx-10\int{x\sin \left( \dfrac{1}{5}x \right)dx}, we get
Let u=10xu=10x and v=5cos(15x)v=5\cos \left( \dfrac{1}{5}x \right)
Then, du=10dxdu=10dx and dv=sin(15x)dv=-\sin \left( \dfrac{1}{5}x \right) .
We can substitute the above values in the formula (3), we get
10xsin(15x)dx=50xcos(15x)50cos(15x)dx-10\int{x\sin \left( \dfrac{1}{5}x \right)dx}=50x\cos \left( \dfrac{1}{5}x \right)-50\int{\cos \left( \dfrac{1}{5}x \right)dx}
We can see that the last integral is trivial, we get
50cos(15x)dx=250sin(15x)-50\int{\cos \left( \dfrac{1}{5}x \right)}dx=-250\sin \left( \dfrac{1}{5}x \right)
Now we can substitute the above value in the integral (3), we get
x2cos(15x)dx=5x2sin(15x)+50xcos(15x)250sin(15x)\int{{{x}^{2}}\cos \left( \dfrac{1}{5}x \right)dx=5{{x}^{2}}\sin \left( \dfrac{1}{5}x \right)+50x\cos \left( \dfrac{1}{5}x \right)-250\sin \left( \dfrac{1}{5}x \right)}
Now we can evaluate the above resulted value using definite integral, we get

& \int\limits_{\dfrac{\pi }{2}}^{\dfrac{5\pi }{2}}{{{x}^{2}}\cos \left( \dfrac{1}{5}x \right)}dx=5{{\left( \dfrac{5\pi }{2} \right)}^{2}}\sin \left( \dfrac{1}{5}\dfrac{5\pi }{2} \right)+50\left( \dfrac{5\pi }{2} \right)\cos \left( \dfrac{1}{5}\dfrac{5\pi }{2} \right)-250\sin \left( \dfrac{1}{5}\dfrac{5\pi }{2} \right) \\\ & \text{ }-\left( 5{{\left( \dfrac{\pi }{2} \right)}^{2}}\sin \left( \dfrac{1}{5}\dfrac{\pi }{2} \right) \right)+50\left( \dfrac{\pi }{2} \right)\cos \left( \dfrac{1}{5}\dfrac{\pi }{2} \right)-250\sin \left. \left( \dfrac{1}{5}\dfrac{\pi }{2} \right) \right) \\\ \end{aligned}$$ We can now simplify the above step, we get $$\int\limits_{\dfrac{\pi }{2}}^{\dfrac{5\pi }{2}}{{{x}^{2}}\cos \left( \dfrac{1}{5}x \right)}dx=8.45459+392.5515-6.853033-0.067644-78.53863+1.37077$$ $$\int\limits_{\dfrac{\pi }{2}}^{\dfrac{5\pi }{2}}{{{x}^{2}}\cos \left( \dfrac{1}{5}x \right)}dx\approx 316.91$$ Therefore, the definite integral value is $$\int\limits_{\dfrac{\pi }{2}}^{\dfrac{5\pi }{2}}{{{x}^{2}}\cos \left( \dfrac{1}{5}x \right)}dx\approx 316.91$$ **Note:** Students make mistakes, while finding each integral one by one using integration by parts, we should concentrate while substituting each integral by parts in the final answer of the integration. We can also use a calculator to find the exact value of the above integral.