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

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

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integral of 1/(2sin(x))

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

\frac{1}{2} ln tan (\frac{x}{2}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

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

= \frac{1}{2} ln tan (\frac{x}{2})

Add a constant to the solution

= \frac{1}{2} ln tan (\frac{x}{2}) + C

4 x^{3} - 48 x^{2} + 6 x + 4

e^{x} \frac{d y}{d x} + x y^{2} = 0

\int cos (x^{2} + 1) d x

y = x, 5 y = x + 6

\frac{d y}{d x} = \frac{4 y}{3 x}

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