Why Is the Coefficient Doubled in the Fourier Sine Series Calculation?

[Nikitin]

Homework Statement

https://wiki.math.ntnu.no/_media/tma4120/2013h/tma4120_h11.pdf

Check out the solution to problem 4b)

My question is: Why do they set $b_n = \frac{2}{\pi} \int_{0}^{\pi}(...)dx$ instead of $b_n = \frac{1}{\pi} \int_{0}^{\pi} (...)dx$?

Ie, why did they multiply the integral with 2? Did they find the Fourier integral to the odd expansion of $sin(x)^2$? Because that's the only way for sin(x)^2 to be defined by solely a sine Fourier expansion?

EDIT: Ooops, posted in wrong section. Please move to calculus and beyond homework forum! sorry!

 

[LCKurtz]

Homework Statement

https://wiki.math.ntnu.no/_media/tma4120/2013h/tma4120_h11.pdf

Check out the solution to problem 4b)

My question is: Why do they set $b_n = \frac{2}{\pi} \int_{0}^{\pi}(...)dx$ instead of $b_n = \frac{1}{\pi} \int_{0}^{\pi} (...)dx$?

Ie, why did they multiply the integral with 2? Did they find the Fourier integral to the odd expansion of $sin(x)^2$? Because that's the only way for sin(x)^2 to be defined by solely a sine Fourier expansion?

Yes, that's exactly it. Since they only have sine functions for the eigenfunction expansion, the FS must represent an odd function so they take the odd extension of $\sin^2(x)$ and use the half range formulas.