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

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

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integral of 1/(tan(θ)sec(θ))

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

ln tan (\frac{θ}{2}) + cos (θ) + C

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

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 ( θ )} d θ - \int sin (θ) d θ

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

\int sin (θ) d θ = - cos (θ)

= ln tan (\frac{θ}{2}) - (- cos (θ))

Simplify

= ln tan (\frac{θ}{2}) + cos (θ)

Add a constant to the solution

= ln tan (\frac{θ}{2}) + cos (θ) + C

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