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

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

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integral of (csc(y))/1

\int \frac{csc ( y )}{1} d y

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

Solution steps

\int \frac{csc ( y )}{1} 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

lo g_{10} (\frac{1}{x})

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

\int \frac{( 2 ln ( x ) ) ^{2}}{4 x} d x

\int \frac{9}{- x ^{2} + 2 x} d x

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