Find the integral : `int 1/(cos 4x) dx`

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The integral `int 1/(cos 4x) dx` has to be determined.

`int 1/(cos 4x) dx`

=> `int (cos 4x)/(cos^2(4x)) dx`

=> `int (cos 4x)/(1 - sin^2 4x)dx`

let sin 4x = y

`dy/dx = 4*cos 4x`

=> `(1/4)*dy = cos 4x dx`

`int (cos 4x)/(1 - sin^2 4x) dx`

=> `(1/4)*int 1/(1- y^2) dy`

=> `(-1/4)*int 1/((y - 1)(y + 1)) dy`

=> `(-1/8)*int ((y + 1) - (y - 1))/((y - 1)(y + 1)) dy`

=> `(-1/8)*int 1/(y - 1) - 1/(y + 1) dy`

=> `(-1/8)*(ln(y - 1) - ln(y + 1)) + C`

=> `(ln(y + 1) - ln(y - 1))/8 + C`

substitute `y = sin 4x`

=> `(ln(sin (4x) + 1) - ln(sin(4x) - 1))/8 + C`

The integral` int 1/(cos 4x) dx` = `(ln(sin (4x) + 1) - ln(sin(4x) - 1))/8 + C`

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