calculus area., where n is a positive integer.

Actually, the two above are just the same, because \(\displaystyle \sin{(ax+b)}=\cos{(ax+b-\frac{\pi}{2})}=\cos{(ax+c)}\), where

\(\displaystyle c=b-\frac{\pi}{2}\).

So, \(\displaystyle I_n=\int x^n\cos{(ax+b)}dx\) is the only one I want to discuss here.

Firstly I give the formula as below:

\(\displaystyle I_n=\int x^n\cos{(ax+b)}dx =\sum_{m=0}^{n}{\frac{\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\sin{(ax+b+\frac{m\pi}{2})}\),

where \(\displaystyle \Gamma(.+1)=(.)!\)

Proof:

It is easy to show that, the integral:

\(\displaystyle \int x^n\e{(ax+b)}dx\)

\(\displaystyle =\sum_{m=0}^{n}{\frac{(-1)^{m}\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\e (ax+b)\)

By using the Euler's formula, \(\displaystyle \cos t=\frac{\e (-it)+\e (it)}{2}\)(\(\displaystyle i\) denotes the imaginary unit):

\(\displaystyle I_n=\int x^n\cos{(ax+b)}dx\)

\(\displaystyle =\frac{1}{2}\int x^n(\e{[-i(ax+b)]}+\e{[i(ax+b)]})dx\)

\(\displaystyle =\frac{1}{2}\{\sum_{m=0}^{n}{\frac{(-1)^{m}\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)(-ia)^{m+1}}\e {[-i(ax+b)]}+\sum_{m=0}^{n}{\frac{(-1)^{m}\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)(ia)^{m+1}}\e {[i(ax+b)]}\}\)

We know,

\(\displaystyle (-1)^m=\e (-im\pi)=\e (im\pi)\)

\(\displaystyle (-i)^{m}=\e (-i\frac{m\pi}{2})\)

\(\displaystyle (i)^{m}=\e (i\frac{m\pi}{2})\),

then,

\(\displaystyle I_n=\frac{1}{2}\sum_{m=0}^{n}{\frac{\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\{\frac{(-1)^{m}\e {[-i(ax+b)]}}{(-i)^{m+1}}+\frac{(-1)^{m}\e {[i(ax+b)]}}{i^{m+1}}\}\)

\(\displaystyle =\frac{1}{2}\sum_{m=0}^{n}{\frac{\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\{\frac{\e (-im\pi)\e {[-i(ax+b)]}}{(-i)\e (-i\frac{m\pi}{2})}+\frac{\e (im\pi)\e {[i(ax+b)]}}{i\e (i\frac{m\pi}{2})}\}\)

\(\displaystyle =\sum_{m=0}^{n}{\frac{\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\{\frac{\e {[i(ax+b+\frac{m\pi}{2})]}-\e {[-i(ax+b+\frac{m\pi}{2})]}}{2i}\}\)

\(\displaystyle =\sum_{m=0}^{n}{\frac{\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\sin{(ax+b+\frac{m\pi}{2})}\).

Q.E.D.