Since
sinx=sum_{n=0}^infty(-1)^n{x^{2n+1}}/{(2n+1)!}sinx=∞∑n=0(−1)nx2n+1(2n+1)!,
we have
sin(sqrt{x})=sum_{n=0}^infty{(-1)^n(sqrt{x})^{2n+1}}/{(2n+1)!}=sum_{n=0}^infty{(-1)^nx^{n+1/2}}/{(2n+1)!}sin(√x)=∞∑n=0(−1)n(√x)2n+1(2n+1)!=∞∑n=0(−1)nxn+12(2n+1)!
Now, consider the integral in question.
int_0^{0.01}sin(sqrt{x})dx=int_0^{0.01}sum_{n=0}^infty{(-1)^nx^{n+1/2}}/{(2n+1)!}dx∫0.010sin(√x)dx=∫0.010∞∑n=0(−1)nxn+12(2n+1)!dx
by integrating term by term,
=sum_{n=0}^infty[{(-1)^nx^{n+3/2}}/{(2n+1)!(n+3/2)}]_0^{0.01}=∞∑n=0⎡⎢⎣(−1)nxn+32(2n+1)!(n+32)⎤⎥⎦0.010
={(0.01)^{3/2}}/{1!cdot3/2}-{(0.01)^{5/2}}/{3!cdot 5/2}+{(0.01)^{7/2}}/{5!cdot 7/2}-cdots=(0.01)321!⋅32−(0.01)523!⋅52+(0.01)725!⋅72−⋯
={2(0.1)^3}/{3cdot 1!}-{2(0.1)^5}/{5cdot 3!}+{2(0.1)^7}/{7cdot 5!}-cdots=2(0.1)33⋅1!−2(0.1)55⋅3!+2(0.1)77⋅5!−⋯
By adding a few terms of the above series, we can approximate the value of the original definite integral.
