What is the integral of (tan(5x)-sec(5x))^2 ?

The integral of (tan(5x)-sec(5x))^2 is -x+2/5 tan(5x)-2/5 sec(5x)+C

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integral of (tan(5x)-sec(5x))^2

\int (tan (5 x) - sec (5 x))^{2} d x

- x + \frac{2}{5} tan (5 x) - \frac{2}{5} sec (5 x) + C

Solution steps

\int (tan (5 x) - sec (5 x))^{2} d x

Expand

(tan (5 x) - sec (5 x))^{2} : tan^{2} (5 x) - 2 tan (5 x) sec (5 x) + sec^{2} (5 x)

= \int tan^{2} (5 x) - 2 tan (5 x) sec (5 x) + sec^{2} (5 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} (5 x) d x - \int 2 tan (5 x) sec (5 x) d x + \int sec^{2} (5 x) d x

\int tan^{2} (5 x) d x = - x + \frac{1}{5} tan (5 x)

\int 2 tan (5 x) sec (5 x) d x = \frac{2}{5} sec (5 x)

\int sec^{2} (5 x) d x = \frac{1}{5} tan (5 x)

= - x + \frac{1}{5} tan (5 x) - \frac{2}{5} sec (5 x) + \frac{1}{5} tan (5 x)

Add similar elements:

\frac{1}{5} tan (5 x) + \frac{1}{5} tan (5 x) = \frac{2}{5} tan (5 x)

= - x + \frac{2}{5} tan (5 x) - \frac{2}{5} sec (5 x)

Add a constant to the solution

= - x + \frac{2}{5} tan (5 x) - \frac{2}{5} sec (5 x) + C

\frac{d}{d x} (2^{x})

\frac{d y}{d x} = y^{2} e^{x} sin (3 x)

\frac{d}{d x} (\frac{2}{x ^{3}} - \frac{1}{x ^{2}})

\int sin (3 t) d t

e^{\frac{1}{x}}

What is the integral of (tan(5x)-sec(5x))^2 ?
The integral of (tan(5x)-sec(5x))^2 is -x+2/5 tan(5x)-2/5 sec(5x)+C