While asking my previous question , I wanted to solve $(1)$ that is stated below this sentence:
$$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{1+\text{L}\cdot\text{C}_2\cdot\text{s}^2}\cdot\mathcal{L}_t\left[\left|\sin\left(\omega t+\theta\right)\right|\right]_{\left(\text{s}\right)}\right]_{\left(t\right)}\tag1$$
Using the answer of #msm, I was able to get (using convolution theorem and where $*$ is the convolution of the two functions):
$$\left\{\frac{1}{\sqrt{\text{L}\cdot\text{C}_2}}\cdot\sin\left(\frac{1}{\sqrt{\text{L}\cdot\text{C}_2}}\cdot t\right)\right\}\space*\space\left|\sin\left(\omega t+\theta\right)\right|\tag2$$
Now, to undo the convolution, I have to solve the integral (posted in red under this sentence):
$$\color{red}{\mathcal{I}=\frac{1}{\sqrt{\text{L}\cdot\text{C}_2}}\cdot\int_0^t\sin\left(\frac{t-\tau}{\sqrt{\text{L}\cdot\text{C}_2}}\right)\cdot\left|\sin\left(\omega\tau+\theta\right)\right|\space\text{d}\tau}\tag3$$
Where I was thinking of expanding function: $\left|\sin\left(\omega\tau+\theta\right)\right|$ into its $\cos$ Fourier series like this:
$$\mathcal{I}=\frac{1}{\sqrt{\text{L}\cdot\text{C}_2}}\cdot\int_0^t\sin\left(\frac{t-\tau}{\sqrt{\text{L}\cdot\text{C}_2}}\right)\cdot\left\{\frac{2}{\pi}-\frac{4}{\pi}\sum_{\text{n}\ge1}\frac{\cos\left(2\text{n}\left(\omega\tau+\theta\right)\right)}{4\text{n}^2-1}\right\}\space\text{d}\tau$$