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Integral-03

Albert

Well-known member
Jan 25, 2013
1,225
evaluate :

$\int_{0}^{2\pi}x^2 cos(nx)\, dx$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

Albert

Well-known member
Jan 25, 2013
1,225
Re: integral-03

perfect (Yes) you got it
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

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

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

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