Using the Parseval’s identity prove that $$\int_{-\pi}^\pi\cos^{4}(x)dx= \frac{3\pi}{4}$$
As far as I know to do this problem we need to find the fourier coefficients of $\cos^{2}(x)$. I am getting $a_{0} = 1$. Also $a_{n}$ is $0$ and $b_{n}$ is $0$. On using the Parseval’s identity I am not getting $\frac{3\pi}{4}$ as the answer
Any correction or help is appreciated. Thanks in Advance
Don't bother evaluating the Fourier series. We know that $\cos^2(x)=\frac12+\frac{\cos(2x)}2=\frac12+\frac{e^{i2x}+e^{-i2x}}4$. So, by Parseval's identity, $\int_{-\pi}^{\pi}\cos^4(x)\,dx=2\pi\left(\frac14+\frac1{16}+\frac1{16}\right)$.