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

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 sin (x) sec (x) d x

\frac{d}{d x} (sin (x) \cdot tan (x))

\frac{d x}{d t} = a x - b x^{2} - 0

\frac{2 x ^{2} + 4 x + 2}{x}

y = 4 x^{3} - 4 x

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