Just like the other one:

The Fourier Series periodic with period 2L, L>0 is defined to be:

(a_0)/2 + sum_{n=1 to infty} [a_n cos(n pi x/L) + b_n sin(n pi x/L)]

where:

a_n = (1/L) integral_{x=-L to L} f(x) cos(n pi x/L) dx

b_n = (1/L) integral_{x=-L to L} f(x) sin(n pi x/L) dx

Now consider f(x) periodic with period pi, where:

f(x) = sin(x), -pi/2<x<pi/2

Now f(x) is odd so all the integrals for the "a" terms are zero. So lets now

consider the "b" terms, here L=pi/2 so:

b_n = (2/pi) integral_{x=-pi/2 to pi/2} sin(x) sin(2 n x) dx

Which I hope I can leave to you.

RonL