Evaluating the Integral $\int_{0}^{2\pi}x^2 cos(nx)\, dx$

[Albert1]

evaluate :

$\int_{0}^{2\pi}x^2 cos(nx)\, dx$

 

[MarkFL]

Re: integral-03

We are given to evaluate:

\(\displaystyle I=\int_0^{2\pi}x^2\cos(nx)\,dx\) where (presumably) \(\displaystyle n\in\mathbb{N}\)

Using integration by parts, we may let:

\(\displaystyle u=x^2\,\therefore\,du=2x\,dx\)

\(\displaystyle dv=\cos(nx)\,dx\,\therefore\,v=\frac{1}{n}\sin(nx)\)

And we have:

\(\displaystyle I=\left.\frac{x^2}{n}\sin(nx) \right|_0^{2\pi}-\frac{2}{n}\int_0^{2\pi} x\sin(nx)\,dx\)

\(\displaystyle I=-\frac{2}{n}\int_0^{2\pi} x\sin(nx)\,dx\)

Using integration by parts again, where:

\(\displaystyle u=x\,\therefore\,du=dx\)

\(\displaystyle dv=\sin(x)\,dx\,\therefore\,v=-\frac{1}{n}\cos(nx)\)

Now we have:

\(\displaystyle I=\frac{2}{n}\left(\left.\frac{x}{n}\cos(nx) \right|_0^{2\pi}+\frac{1}{n}\int_0^{2\pi}\cos(nx)\,dx \right)\)

\(\displaystyle I=\frac{2}{n}\left(\frac{2\pi}{n}+\left.\frac{1}{n^2}\sin(nx) \right|_0^{2\pi} \right)\)

\(\displaystyle I=\frac{4\pi}{n^2}\)

Thus, we may state:

\(\displaystyle \int_0^{2\pi}x^2\cos(nx)\,dx=\frac{4\pi}{n^2}\)

 

[Albert1]

Re: integral-03

perfect (Yes) you got it

 

[MarkFL]

Re: integral-03

Let's generalize a little...

We are given to evaluate:

\(\displaystyle I=\int_0^{2\pi}x^2\cos\left(nx+m\frac{\pi}{2} \right)\,dx\) where (presumably) \(\displaystyle n\in\mathbb{N},\,m\in\{0,1,2,3\}\)

Using integration by parts, we may let:

\(\displaystyle u=x^2\,\therefore\,du=2x\,dx\)

\(\displaystyle dv=\cos\left(nx+m\frac{\pi}{2} \right)\,dx\,\therefore\,v=\frac{1}{n}\sin\left(nx+m\frac{\pi}{2} \right)\)

And we have:

\(\displaystyle I=\left.\frac{x^2}{n}\sin\left(nx+m\frac{\pi}{2} \right) \right|_0^{2\pi}-\frac{2}{n}\int_0^{2\pi} x\sin\left(nx+m\frac{\pi}{2} \right)\,dx\)

\(\displaystyle I=\frac{4\pi^2}{n}\sin\left(m\frac{\pi}{2} \right)-\frac{2}{n}\int_0^{2\pi} x\sin\left(nx+m\frac{\pi}{2} \right)\,dx\)

Using integration by parts again, where:

\(\displaystyle u=x\,\therefore\,du=dx\)

\(\displaystyle dv=\sin\left(nx+m\frac{\pi}{2} \right)\,dx\,\therefore\,v=-\frac{1}{n}\cos\left(nx+m\frac{\pi}{2} \right)\)

Now we have:

\(\displaystyle I=\frac{4\pi^2}{n}\sin\left(m\frac{\pi}{2} \right)+\frac{2}{n}\left( \left.\frac{x}{n}\cos\left(nx+m\frac{\pi}{2} \right) \right|_0^{2\pi}+\frac{1}{n}\int_0^{2\pi}\cos\left(nx+m\frac{\pi}{2} \right)\,dx \right)\)

\(\displaystyle I=\frac{4\pi^2}{n}\sin\left(m\frac{\pi}{2} \right)+ \frac{2}{n}\left(\frac{2\pi}{n}\cos\left(m \frac{\pi}{2} \right)+\left.\frac{1}{n^2}\sin\left(nx+m\frac{\pi}{2} \right) \right|_0^{2\pi} \right)\)

\(\displaystyle I=\frac{4\pi^2}{n}\sin\left(m\frac{\pi}{2} \right)+\frac{4\pi}{n^2}\cos\left(m\frac{\pi}{2} \right)\)

Thus, we may state:

\(\displaystyle m=0\implies\int_0^{2\pi}x^2\cos\left(nx \right)\,dx=\frac{4\pi}{n^2}\)

\(\displaystyle m=1\implies-\int_0^{2\pi}x^2\sin\left(nx \right)\,dx=\frac{4\pi^2}{n}\)

\(\displaystyle m=2\implies-\int_0^{2\pi}x^2\cos\left(nx \right)\,dx=-\frac{4\pi}{n^2}\)

\(\displaystyle m=3\implies\int_0^{2\pi}x^2\sin\left(nx \right)\,dx=-\frac{4\pi^2}{n}\)

 

[CellCoree]

Evaluating this integral requires the use of integration techniques, such as integration by parts or substitution. However, based on the given limits of integration and the integrand, it appears that this integral will result in a periodic function. This is because the integrand contains the cosine function, which has a period of 2π. Therefore, the integral will have the same value for any interval of 2π. As a result, the integral can be evaluated by simply finding the value of the integral for one interval and multiplying it by the number of intervals, which in this case is 2π. In summary, the integral $\int_{0}^{2\pi}x^2 cos(nx)\, dx$ can be evaluated by finding the value of the integral for one interval of 2π and multiplying it by 2π.