The antiderivative $f\left(x\right)=\frac{1}{3+5\sin x+3\cos x}$ whose graph passes through the point $(0,0)$ is

$f\left(x\right)=\frac{1}{3+5\sin x+3\cos x}$
We know that, $\sin x=\frac{2{\tan }^x/2}{1+{\tan }^{2}x/2}$ and $\cos x=\frac{1-{\tan }^{2}x/2}{1+{\tan }^{2}x/2}$
$I=\int f\left(x\right)=\int \frac{1}{3+5·2{\tan }^{x/2}}+3·\frac{1-{\tan }^{2}x/2}{1+{\tan }^{2}x/2}$
$I=\int \frac{1+{\tan }^{2}x/2}{10\tan x/2+6}=\int \frac{{\sec }^{2}x/2}{10{\tan }^x/2+6}dx$
Let $\tan x/2=u$
${\sec }^{2}x/2·\frac{1}{2}dx=du$

$I=\int \frac{2du}{10u+6}=\frac{\ln (5u+3)}{5}+c$
$\int f^{\text{'}}\left(x\right)=\frac{1}{5}\ln (5\tan x/2+3)+c$
$\int f\left(0\right)=0\Rightarrow c=-\frac{\ln 3}{5}$
$\Rightarrow \int f\left(x\right).dx=\frac{1}{5}\left(\log \left|1+\frac{5}{3}\cot x/2\right|\right)$