`int 1/(cos2x) dx`

We know that `secx = 1/(cosx) `

==> `int 1/(cos2x) dx = int sec(2x) dx`

Let `u= 2x ==gt du = 2 dx `

`==gt int sec(2x) dx = int sec(u) (du)/2 = (1/2) int sec(u) du`

Now we know that:

`int sec(u) du = ln (secu...

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`int 1/(cos2x) dx`

We know that `secx = 1/(cosx) `

==> `int 1/(cos2x) dx = int sec(2x) dx`

Let `u= 2x ==gt du = 2 dx `

`==gt int sec(2x) dx = int sec(u) (du)/2 = (1/2) int sec(u) du`

Now we know that:

`int sec(u) du = ln (secu + tanu)+ C `

`==gt int 1/(cos2x) dx = (1/2) ln (sec(2x) + tan(2x) )+ C`