What is the integral of cot(x)sec(x) ?

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

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integral of cot(x)sec(x)

\int cot (x) sec (x) d x

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

Solution steps

\int cot (x) sec (x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = tan (\frac{x}{2})

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

Add a constant to the solution

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

\int \frac{x + 6}{( x + 1 ) ^{2}} d x

\frac{\partial ^{2}}{\partial r \partial θ} (r (θ) cos (θ))

x \to 0 lim (\frac{x ^{2} + 1}{x - 1})

(x + y + 1) d x - (y - x + 3) d y = 0

x \to - 1 lim (7 ∣ x + 1 ∣ - 4)

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