What is the integral of 1/(sin(u)) ?

The integral of 1/(sin(u)) is ln|tan(u/2)|+C

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integral of 1/(sin(u))

\int \frac{1}{sin ( u )} d u

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

Solution steps

\int \frac{1}{sin ( u )} d u

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= ln ∣ v ∣

Substitute back

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

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

Add a constant to the solution

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

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What is the integral of 1/(sin(u)) ?
The integral of 1/(sin(u)) is ln|tan(u/2)|+C