Correct choice is (b) 4/π sin x
To elaborate: We know that the Fourier series is given by a0 + \(∑_{n=1}^∞\) (an cos nx + bn sin nx)
This can be also written as a0 + a1cos x + b1 sin x + a2 cos 2x + b2 sin 2x ……
Here, c0 = a0 = 0
So, cn = bn = \(2∫_0^t f(t).sinωt \,dt\)
∴ c1 = b1 = \(2∫_0^t sinωt \,dt\)
= \(– \frac{2}{ω} [cosωt]_0^t\)
= \(-\frac{2}{π}\)[-1-1] = 4/π
And a1 = \(2∫_0^t f(t).cosωt \,dt\)
= \(\frac{2}{ω}[sinωt]_0^t\)
= 0
So, the Fourier series can be written as 0 + 0.cos x + 4/π sin x
= 4/π sin x.