Originally Posted by

**Mathstud28** since sin(x)≡$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ woulding $\displaystyle sin(x)^{m}$ or $\displaystyle [sin(x)]^{m}$

be the same as $\displaystyle \bigg[\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\bigg]^{m}$ which is equal to $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m}}{[(2n+1)!]^{m}}$ so $\displaystyle \int\sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m}}{[(2n+1)!]^{m}},dx$=$\displaystyle \sum_{n=0}^{\infty}\int\frac{(-1)^{mn}x^{2mn+mm}}{[(2n+1)!]^{m}},dx$ which is equal to $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{mn}}{[(2n+1)!]^{m}}\int[x^{2nm+m}],dx$=$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m+1}}{[(2n+1)!]^{m}(2mn+m+1)}$