∫ tan ^3 (2x)sec (2x) dx give a hint to solve it

please shows the beginning steps too

For the integrand tan^3(2 x) sec(2 x), substitute u = 2 x and du = 2 dx: =1/2 essential tan^3(u) sec(u) du 

For the integrand tan^3(u) sec(u), utilize the trigonometric personality tan^2(u) = sec^2(u) - 1: 

=1/2 vital tan(u) sec(u) (sec^2(u) - 1) du 

For the integrand tan(u) sec(u) (sec^2(u) - 1), substitute s = sec(u) and ds = tan(u) sec(u) du: =1/2 vital (s^2 - 1) ds 

Incorporating term by term and figuring out constants: 

=1/2 vital s^2 ds - 1/2 vital 1 ds 

vital of s^2 is s^3/3: 

=s^3/6 - 1/2 vital 1 ds 

vital of 1 is s:=s^3/6 - s/2 + consistent 

Substitute back for s = sec(u): 

=(sec^3(u))/6 - (sec(u))/2 + consistent 

Substitute back for u = 2 x: 

=1/6 sec^3(2 x) - 1/2 sec(2 x) + consistent 

Which is equivalent to: 

=> 1/6 sec(2 x) (sec^2(2 x) - 3) + c.
 

Cheers!!

Regards,

Vikas (B. Tech. 4^th year
Thapar University)