`(dy)/(dx) = 1/cos(2x)`

`y = int1/cos(2x)dx`

Using sec(2x);

`y=intsec(2x)dx`

`y = 1/2ln|sec(2x)+tan(2x)|+c`

Now let us see how we can do this without using sec(2x);

`y = int1/cos(2x)dx`

Let's write cos(2x) = cos^2(x)-sin^2(x)

Then,

`y = int1/(cos^2(x)-sin^2(x))dx`

Dividing both numerator and denominator by cos^2(x).

`y = int(1/(cos^2(x))/((cos^2(x)-sin^2(x))/cos^2(x)))dx`

`y = int(sec^2(x))/(1-tan^2(x))dx`

but we know, d(tan^2(x)) = sec^2(x)

Therefore,

`y = int1/(1-tan^2(x))d(tan(x))`

Using partial fractions,

`y =(1/2) int(1/(1-tan(x))+1/(1+tan(x)))d(tan(x))`

Integrating wrt tan(x) would give,

`y = (1/2)(-ln(1-tan(x))+ln(1+tan(x)))+ c`

`y = (1/2)ln((1+tan(x))/(1-tan(x)))+ c`

`y = (1/2)ln((cos(x)+sin(x))/(cos(x)-sin(x)))+ c`

Multiplying numerator and denominator inside ln by (cos(x)+sin(x))

`y = (1/2)ln((cos(x)+sin(x))^2/(cos^2(x)-sin^2(x)))+ c`

`y = (1/2)ln((1+sin(2x))/(cos(2x)))+ c`

`y = (1/2)ln(sec(2x)+tan(2x))+ c`

We get the same answer as above.