What is the integral of 1/(sin^2(y)csc(y)) ?

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

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integral of 1/(sin^2(y)csc(y))

\int \frac{1}{sin ^{2} ( y ) csc ( y )} d y

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

Solution steps

\int \frac{1}{sin ^{2} ( y ) csc ( y )} d y

Simplify

\frac{1}{sin ^{2} ( y ) csc ( y )} : csc (y)

= \int csc (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

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