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

The integral of (sec(x)-cot(x))^2 is tan(x)-2ln|tan(x/2)|-x-cot(x)+C

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integral of (sec(x)-cot(x))^2

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

tan (x) - 2 ln tan (\frac{x}{2}) - x - cot (x) + C

Solution steps

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

Expand

(sec (x) - cot (x))^{2} : sec^{2} (x) - 2 sec (x) cot (x) + cot^{2} (x)

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

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

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

\int cot^{2} (x) d x = - x - cot (x)

= tan (x) - 2 ln tan (\frac{x}{2}) - x - cot (x)

Add a constant to the solution

= tan (x) - 2 ln tan (\frac{x}{2}) - x - cot (x) + C

\frac{x ^{9} + 7}{x ^{9}}

\frac{d}{d x} (1 0^{(x^{2} - 1)})

\int \frac{15 e ^{5 x}}{e ^{5 x} + 2} d x

f (x) = - 4 x^{2} + 7 x - 14, at x = - 2

\frac{\partial}{\partial x} (x^{c o s (x)})

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