What is the integral of (sec(2x)-csc(x/2)) ?

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

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integral of (sec(2x)-csc(x/2))

\int (sec (2 x) - csc (\frac{x}{2})) d x

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

Solution steps

\int sec (2 x) - csc (\frac{x}{2}) d x

Rewrite using trig identities

= \int \frac{sin ( \frac{x}{2} ) - cos ( 2 x )}{cos ( 2 x ) sin ( \frac{x}{2} )} d x

Expand

\frac{sin ( \frac{x}{2} ) - cos ( 2 x )}{cos ( 2 x ) sin ( \frac{x}{2} )} : \frac{1}{cos ( 2 x )} - \frac{1}{sin ( \frac{x}{2} )}

= \int \frac{1}{cos ( 2 x )} - \frac{1}{sin ( \frac{x}{2} )} 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 \frac{1}{cos ( 2 x )} d x - \int \frac{1}{sin ( \frac{x}{2} )} d x

\int \frac{1}{cos ( 2 x )} d x = \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣

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

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

Add a constant to the solution

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

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