What is the integral of 1/(-sin(x)) ?

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

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integral of 1/(-sin(x))

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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

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