How to find properties of time varying voltage?
Do the integrals with \$\small sin(5\omega t)\$ and \$\small cos(5\omega t)\$ replacing \$\small sin(\omega t)\$ and \$\small cos(\omega t)\$, respectively. The amplitude of the \$\small 5^{th}\$ harmonic is then: \$\small \sqrt{a_5^2+b_5^2}\$ and the phase angle is: \$ \small\phi_n= arctan\large \left (\frac{b_5}{a_5}\right)\$

In general, the amplitude and phase angle of the \$\small n^{th}\$ harmonic are:
 $$\small A_n=\sqrt{a_n^2+b_n^2}$$ and: $$\small \phi_n = \small arctan \left (\frac{b_n}{a_n}\right)$$.

To see why this is so, consider the simple case: \$\small y(t)= A sin(\omega t+\phi)\$, which may be written: $$\small y(t)= Acos(\phi)\:sin(\omega t)+Asin(\phi)\:cos(\omega t)$$

Now, doing the Fourier integral gives:
$$\small a_1= Acos(\phi) $$ 
$$\small b_1= Asin(\phi) $$
Hence:$$ \small A= \sqrt{a_1^2+b_1^2}$$
$$ \small tan(\phi)= \left (\frac{b_1}{a_1}\right)$$

We can easily expand this analysis to the \$\small n^{th}\$ harmonic giving:
$$\small A_n=\sqrt{a_n^2+b_n^2}$$
$$ \small tan(\phi_n)= \left (\frac{b_n}{a_n}\right)$$