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

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

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integral of 1/(sin(y/2)cos(y/2))

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

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

Solution steps

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

Rewrite using trig identities

= \int \frac{2}{sin ( y )} d y

Take the constant out:

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

= 2 \cdot \int \frac{1}{sin ( y )} d y

Apply u-substitution

: \int \frac{1}{u} d u

= 2 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 ln ∣ u ∣

Substitute back

u = tan (\frac{y}{2})

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

Add a constant to the solution

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

\int \frac{sin ( 2 x )}{1 - cos ( 2 x )} d x

\int sin (2 x) e^{5 x} d x

\int \frac{x ^{2} + 1}{x ^{2} - 3 x + 2} d x

\int \frac{ln ( x ) + 1}{x} d x

\int \frac{2 x ^{2} - 1}{x ( 1 - x ^{2} )} d x

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