What is the integral of-1/(sin(θ)) ?

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

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integral of-1/(sin(θ))

\int - \frac{1}{sin ( θ )} d θ

- ln tan (\frac{θ}{2}) + C

Solution steps

\int - \frac{1}{sin ( θ )} d θ

Take the constant out:

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

= - \int \frac{1}{sin ( θ )} d θ

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{θ}{2})

= - ln tan (\frac{θ}{2})

Add a constant to the solution

= - ln tan (\frac{θ}{2}) + C

y^{^{'}} + t^{2} y = t^{2}

\frac{d}{d x} (9 x e^{- \frac{1}{y ^{3}}})

\int \frac{1}{6 - x ^{2}} d x

\frac{d}{d x} (\frac{5}{7 x ^{2}})

3 x + x

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