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

Rewrite using trig identities

= \int sec^{2} (x) + 2 sec (x) cot (x) - 1 + csc^{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 1 d x + \int csc^{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 1 d x = x

\int csc^{2} (x) d 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{d}{d x} ((6 x^{2} + 5) (3 x + x))

\frac{x + 2}{x ^{2} - 4}

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

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

\frac{1}{s + 4}

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