How to calculate the integral of square of Fourier series.

Question

In the proof of Bessel's inequality, I need to expand the following: $$ \int_{-\pi}^{\pi}{\left[\sum_{v=1}^{n}\left( a_v\cos{vx}+b_v\sin{vx} \right)\right]^2dx} $$ I do it like this: $$ =\int_{-\pi}^{\pi}{\sum_{v=1}^{n}{ \left( a_v\cos{vx}+b_v\sin{vx} \right)^2}dx} +\int_{-\pi}^{\pi}{ \sum_{n=m+1}^{n}{ \sum_{m=1}^{n-1}{ 2\left( a_m\cos{mx}+b_m\sin{mx} \right) \left( a_n\cos{nx}+b_n\sin{nx} \right) }}dx} \\ =\sum_{v=1}^{n}{\int_{-\pi}^{\pi}{ \left( a_v^2\cos^2{vx}+b_v^2\sin^2{vx} +2a_vb_v\cos{vx}\sin{vx} \right)}dx}\\ +\sum_{n=m+1}^{n}{\sum_{m=1}^{n-1}{\int_{-\pi}^{\pi}{ 2a_ma_n\cos{mx}\cos{nx}}dx}}\\ +\sum_{n=m+1}^{n}{\sum_{m=1}^{n-1}{\int_{-\pi}^{\pi}{ 2a_mb_n\sin{nx}\cos{mx}}dx}}\\ +\sum_{n=m+1}^{n}{\sum_{m=1}^{n-1}{\int_{-\pi}^{\pi}{ 2a_nb_m\sin{mx}\cos{nx}}dx}}\\ +\sum_{n=m+1}^{n}{\sum_{m=1}^{n-1}{\int_{-\pi}^{\pi}{ 2a_nb_m\sin{mx}\cos{nx}}dx}}\\ $$ Did I do the wrong way? Or what should I do next?