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

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

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integral of-1/(sin(y))

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

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

Solution steps

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

Take the constant out:

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

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

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{y}{2})

= - ln tan (\frac{y}{2})

Add a constant to the solution

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

\int e^{- 21 x} d x

\int \frac{1}{x ^{4} - x ^{3}} d x

cos (3 x)

\frac{d}{d x} (\frac{10}{ln ( x )})

t^{2} \frac{d y}{d t} + y = 0, y (7) = - 4

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