What is the integral of 1/(tan(x)sec(x)) ?

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

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integral of 1/(tan(x)sec(x))

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

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

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

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

\int sin (x) d x = - cos (x)

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

Simplify

= ln tan (\frac{x}{2}) + cos (x)

Add a constant to the solution

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

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