Evaluate the following integrals:

in this integral we are going to put the value of $\sin (x)$ in terms of $\tan (x / 2)$ -

$\begin{aligned} I &=\int \frac{2 d t}{5+5 t^{2}-8 t} \\ I &=\frac{2}{5} \int \frac{1}{\left(t-\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}} d t \end{aligned}$

By applying the formula of $1 /\left(x^{2}+a^{2}\right)$ in above equation yields the integral-

$I=\frac{2}{5} \cdot \frac{1}{\frac{3}{5}} \cdot \tan ^{-1} \frac{\left(t-\frac{4}{5}\right)}{\left(\frac{3}{5}\right)}$

$I=\frac{2}{3} \cdot \tan ^{-1} \frac{5 t-4}{3}$

By putting the value of $t$ in above equation,

$I=\frac{2}{3} \cdot \tan ^{-1}\left(\frac{5}{3} \tan \frac{x}{2}-\frac{4}{3}\right)$