`int sec^2(2 theta) d theta` Evaluate the indefinite integral.

You need to use the following substitution  `2theta = t` , such that:

`2theta = t=>2d theta = dt => d theta = (dt)/2`

`int sec^2(2 theta) d theta = (1/2)*int sec^2 t dt`

`(1/2)*int sec^2 t dt =(1/2)*tan t + c`

Replacing back `2theta` for t yields:

`int sec^2(2 theta) d theta = (1/2)*tan(2theta) + c`

Hence, evaluating the indefinite integral, yields `int sec^2(2 theta) d theta = (1/2)*tan(2theta) + c`