Integral of $\sin(x)$ using power series.

Question

$\displaystyle \int_{0}^{1} \sin(x) \, dx$

$\sin(x) = \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$

Integrating this from $0 \to 1$

On the RHS we get

$\displaystyle (-)\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!} $

but I can't find a way to sum this series....

Thanks!

Answer 1

Well, you could get the power series for what you "know" the correct integral should be and compare ... or, on the other hand, you could note that probably your calculator/favorite CAS uses something along the lines of power series to approximate the value of $\cos(x)$...

Answer 2

The power series has an infinite radius, amongst other things it allows you to switch integral and sum:

$ I = \displaystyle \int_{0}^{1} \sin(x) \, dx = \int_0^1 (\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!})dx = \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)!}\int_0^1x^{2n+1}dx =\sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+2)!}$

Now if you want to evaluate this... Well: $ I = [-cos(x)]_0^1 = 1-cos(1) = \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+2)!} $