What is the integral of (tan(4x)-sec(4x))^2 ?

The integral of (tan(4x)-sec(4x))^2 is -x+1/2 tan(4x)-1/2 sec(4x)+C

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integral of (tan(4x)-sec(4x))^2

\int (tan (4 x) - sec (4 x))^{2} d x

- x + \frac{1}{2} tan (4 x) - \frac{1}{2} sec (4 x) + C

Solution steps

\int (tan (4 x) - sec (4 x))^{2} d x

Expand

(tan (4 x) - sec (4 x))^{2} : tan^{2} (4 x) - 2 tan (4 x) sec (4 x) + sec^{2} (4 x)

= \int tan^{2} (4 x) - 2 tan (4 x) sec (4 x) + sec^{2} (4 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int tan^{2} (4 x) d x - \int 2 tan (4 x) sec (4 x) d x + \int sec^{2} (4 x) d x

\int tan^{2} (4 x) d x = - x + \frac{1}{4} tan (4 x)

\int 2 tan (4 x) sec (4 x) d x = \frac{1}{2} sec (4 x)

\int sec^{2} (4 x) d x = \frac{1}{4} tan (4 x)

= - x + \frac{1}{4} tan (4 x) - \frac{1}{2} sec (4 x) + \frac{1}{4} tan (4 x)

Add similar elements:

\frac{1}{4} tan (4 x) + \frac{1}{4} tan (4 x) = \frac{1}{2} tan (4 x)

= - x + \frac{1}{2} tan (4 x) - \frac{1}{2} sec (4 x)

Add a constant to the solution

= - x + \frac{1}{2} tan (4 x) - \frac{1}{2} sec (4 x) + C

What is the integral of (tan(4x)-sec(4x))^2 ?
The integral of (tan(4x)-sec(4x))^2 is -x+1/2 tan(4x)-1/2 sec(4x)+C